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How do I plot sagitta versus arc length.
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I'm working on a curvature sensor, but I'm finding it hard to find the equation for my h (sagitta) as a function of my arc length.
In particular, I want to solve $epsilon=frac{Arc Length - Chord Length}{Chord Length}$ for the sagitta, knowing the Chord Length.
So the first step is to get the Arc Length as a function of Chord Length and sagitta, which I think I can do. The second step would be to plug it in the previous equation and solve it for h (as a function of Chord Length and $epsilon$). Is that possible?
Thanks
geometry circle
asked Jun 5 '15 at 19:44
Luis CostaLuis Costa
1
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add a comment |
$begingroup$
I'm working on a curvature sensor, but I'm finding it hard to find the equation for my h (sagitta) as a function of my arc length.
In particular, I want to solve $epsilon=frac{Arc Length - Chord Length}{Chord Length}$ for the sagitta, knowing the Chord Length.
So the first step is to get the Arc Length as a function of Chord Length and sagitta, which I think I can do. The second step would be to plug it in the previous equation and solve it for h (as a function of Chord Length and $epsilon$). Is that possible?
Thanks
geometry circle
asked Jun 5 '15 at 19:44
Luis CostaLuis Costa
1
$endgroup$
add a comment |
$begingroup$
I'm working on a curvature sensor, but I'm finding it hard to find the equation for my h (sagitta) as a function of my arc length.
In particular, I want to solve $epsilon=frac{Arc Length - Chord Length}{Chord Length}$ for the sagitta, knowing the Chord Length.
So the first step is to get the Arc Length as a function of Chord Length and sagitta, which I think I can do. The second step would be to plug it in the previous equation and solve it for h (as a function of Chord Length and $epsilon$). Is that possible?
Thanks
geometry circle
asked Jun 5 '15 at 19:44
Luis CostaLuis Costa
1
$endgroup$
I'm working on a curvature sensor, but I'm finding it hard to find the equation for my h (sagitta) as a function of my arc length.
In particular, I want to solve $epsilon=frac{Arc Length - Chord Length}{Chord Length}$ for the sagitta, knowing the Chord Length.
So the first step is to get the Arc Length as a function of Chord Length and sagitta, which I think I can do. The second step would be to plug it in the previous equation and solve it for h (as a function of Chord Length and $epsilon$). Is that possible?
Thanks
geometry circle
geometry circle
asked Jun 5 '15 at 19:44
Luis CostaLuis Costa
1
asked Jun 5 '15 at 19:44
Luis CostaLuis Costa
1
asked Jun 5 '15 at 19:44
Luis CostaLuis Costa
1
asked Jun 5 '15 at 19:44
Luis CostaLuis Costa
1
asked Jun 5 '15 at 19:44
Luis CostaLuis Costa
1
1
add a comment |
add a comment |
2 Answers
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Hint:
If the arc lenght is $r theta$ (where $theta$ is the angle subtending the arc $AB$) we have:
For the cord: $AB=2rsin (theta/2)$
For the sagitta: $s= r(1-cos(theta /2)$)
Can you prove these results using simple trigonometry on the circle of radius $r$?
If not see here.
answered Jun 5 '15 at 19:56
Emilio NovatiEmilio Novati
51.8k43474
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The first one seems just like the definition of a Sine (Since AB/2 is one of the short sides of the triangle, and r-s is the other one). In the sagitta case, it's basically just the radius minus the length from center of the circle to the chord (r-s), so you get the sagitta. That's not exactly where my issue lies. I know AB, and I know h at any given time. So here's what I have : I know that $ R = frac{(c/2)^2 + h^2}{2h} $. My chord length is known (calling it c). So, my arc length will be $L = frac{(c/2)^2 + h^2}{2h} times arcsin{ frac{2hc}{h^2+c^2}}$. I can't solve this for h.
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– Luis Costa
Jun 5 '15 at 20:20
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sorry h = sagitta.
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– Luis Costa
Jun 5 '15 at 20:24
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maybe that I don't understand your question. But, if you know the cord $c$ and the sagitta $h$ you can find the radius $r$ solving the equation: $r^2=(c/2)^2+(r-h)^2$ than find the arc $theta$ using using the equation that gives the cord ( or the sagitta).
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– Emilio Novati
Jun 5 '15 at 20:39
add a comment |
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NOTE: SEE BELOW FOR A MAJOR UPDATE BEFORE YOU READ THIS FIRST PART.
Using several equation solver tools, I was unable to converge on an exact solution. Essentially, the equation $L=arcsin(frac{c}{h+frac{c^2}{4h}}*(h+frac{c^2}{4h}))$ has to be arranged to solve for $h$, which broke every solver tool I had (where $c$ is chord length, $h$ is height of segment or sagitta, and $L$ is arc length). Instead, I offer you a $R^2 = 0.9999$ solution from graphed data points.
I popped open Solidworks, a CAD program, which uses an iterative solver to determine this geometry, and I drew a circular segment, constraining at the chord length and arc length.
Image: Screenshot of segment showing arc length of 1.25 in Solidworks
Making sure the cord length was equal to $1$, I changed the arc length little by little and recorded the output height. This gave me the following table of values from a $L/c$ ratio of $1.01$ to $pi/2$:
x = L/c y = h/c
1.01 0.06132899
1.015 0.07516845
1.02 0.08686174
1.03 0.10654164
1.04 0.12320591
1.05 0.13795167
1.1 0.19652075
1.15 0.24241721
1.2 0.28189519
1.25 0.31735497
1.3 0.3500165
1.35 0.38059796
1.4 0.40956308
1.45 0.43723107
1.5 0.46383222
1.55 0.48953882
1.56 0.49458502
1.570796327 0.5
Here is the 5th order quadratic equation driven from that data using mycurvefit.com:
$y = -130.1039 + 489.8105*x - 737.3181*x^2 + 554.5809*x^3 - 208.0569*x^4 + ...$ $31.13006*x^5$
All you do is pop in the ratio $L/c$ for $x$ and then you get $y$, the ratio $h/c$. Since the chord length is known, and you have the ratio $h/c$, you can solve for the height by multiplying the ratio of $h/c$ by the chord length.
...but I kept going...
I wondered if someone reading this might have an application for the inverse segment. Things get less complicated as the ratio of $L/c$ approaches infinity.
Here is the data from $x = pi/2$ to $x = pi$:
x = L/c y = h/c
1.570796327 0.5
1.58 0.50458986
1.59 0.50955019
1.6 0.51448353
1.7 0.56248484
1.8 0.60845528
1.9 0.65281718
2 0.6958726
2.1 0.73784553
2.2 0.778907
2.3 0.81919066
2.4 0.85880306
2.5 0.89783052
2.6 0.93634401
2.7 0.97440257
2.8 1.01205584
2.9 1.04934598
3 1.08630908
3.14159265 1.13814656
This has a polynomial curve fit as follows:
$y = -0.5308871 + 0.8723522*x - 0.1662753*x^2 + 0.01838478*x^3$
Here is the data from $pi$ to $10$:
x = L/c y = h/c
3.14159265 1.13814656
3.25 1.17748049
3.5 1.26717904
3.75 1.35568642
4 1.44321347
4.5 1.61593815
5 1.78627783
5.5 1.9548322
6 2.12201017
7 2.45331872
8 2.78172909
9 3.10812811
10 3.43306996
Which has a polynomial curve fit as follows:
$y = 0.02093882 + 0.364244*x - 0.002337369*x^2$
The relationship between $L/c$ and h/c becomes very linear after this point, as $R^2$ values still are $0.9999$ or better just using a linear equation. Here is the data from $x = 10$ to $100$:
x = L/c y = h/c
10 3.43306996
15 5.04548886
20 6.64810187
25 8.24652077
50 16.21859143
75 24.1813035
100 32.14157177
The linear regression for this curve fit is as follows:
$y = 0.3188628*x + 0.2643024$
Beyond this point, I am not sure if the iterative solver is becoming unreliable due to rounding errors, so use the next information with a grain of salt as it approaches the capabilities of the software. Here is the data from $100$ to $10^{10}$:
x = L/c y = h/c
100 32.14157177
1000 318.627412
100000 31831.30692
10000000000 3183098862
The linear regression is as follows:
$y = 0.3183099*x + 0.3154708$
Despite the growing value of the intercept, it is relatively insignificant in light of the magnitude of the input or output and is probably just a growing rounding error as the software approaches its limitations in the iterative solver. Therefore, beyond $10^{10}$, you can simply use $m = pi^{-1}$ and $b = 0$ in the linear regression formula $y = m*x + b$:
$y = x/pi$
Or in other words, as $x$ approaches infinity, $m$ approaches the inverse of $pi$ and $b$ approaches zero. In practical terms, the smaller the chord is, the closer the sagitta becomes the diameter. I think that passes the smell test.
See this link to download an Excel file solver:
https://drive.google.com/file/d/1er5kIRyL6UahqqAAjlKTEVVhOeFyDZNd/view?usp=sharing
Download it if you want to use it. Don't bother editting or opening in Google Docs. I used features only available in Excel. Once open in Excel, just enter in the Arc and Chord lengths into cells B2 and B3, respectively, and out pops the Sigatta in cell B8.
MAJOR UPDATE
I have since found an additional way to solve this problem thanks to the following resources from 1728.org:
Circle Segment Calculator: http://www.1728.org/circsect.htm
Explanation of Newton's Method: http://www.1728.org/newton.htm
Explained on the second link, I was able to solve for the arc angle using the Newton-Raphson Method of Iterative solving. Usually, the solution converges within 8 decimal places within 4 iterations. However, if the guess is far too great or small, the method fails to find a valid solution (and breaks the solver for the first link). To prevent this, I created a method by which to estimate the initial guess for the half arc angle, mathematically.
$X_{Guess} = pi*(tanh((ln(L/c-1)+0.646543492637008)*0.373118408093598)+1)$
WHERE:
$X_{Guess}$ is an approximation for half the arc angle in radians.
$tanh()$ and $ln()$ are trigonometric and logarithmic functions, respectively.
$L$ is the arc length.
$c$ is the cord length.
The $R^2$ value is about $0.999005$, so this gives a pretty close guess to the arc angle. The largest error I have noticed on my chart is an error value of $2.815% $ as percentage of difference ($X_{Data}$ minus $X_{Guess}$) divided by $pi$. Because it is so close, you can use this as an input to solve for a much closer solution using the Newton-Raphson method as follows:
$f(X_N) = sin(X_N) - c*X_N/L = 0$
$f'(X_N) = cos(X_N) - c/L = 0$
$X_{N+1} = X_N - f(X)/f'(X)$
WHERE:
$f(X_N)$ is an equation of equality given the angle and the chord and arc lengths.
$sin()$ and $cos()$ are trigonometric functions.
$c$ is cord length.
$L$ is the arc length.
$f'(X_N)$ is the derivative of $f(X_N)$.
$X_{N+1}$ is the next iteration of X to solve using the Newton-Raphson Method.
Using the $X_{Guess}$ from the first equation, input it into those three equations to find the next $X_N$ and so on until the $X_N$'s stop changing. For instance, using $L=20$ and $c=18$...
$X_{Guess} = pi*(tanh((ln(20/18-1)+0.646543492637008)*0.373118408093598)+1) = 0.75140903$
$f(X_{Guess}) = sin(0.75140903) - 18*0.75140903/20 = 0.00640093$
$f'(X_{Guess}) = cos(0.75140903) - 18/20 = 0.75140903$
$X_1 = 0.75140903 - 0.00640093/0.75140903 = 0.74289047$
Inputting $X_1$ in place of $X_{Guess}$, $X_2$ in place of $X_1$ and so on...
$X_2 = 0.79071296$
$X_3 = 0.78671238$
$X_4 = 0.78668307$
$X_5 = 0.78668307$
And there we stop, since the solution has converged on 8 decimal places.
Error from guess as a difference ($X_5$ minus $X_{Guess}$) divided by $pi$:
$(0.75140903 - 0.78668307)/pi = -1.123% $
Which, with a 100 FT of material, that 1% could make a difference. Furthermore, this method is arguably more accurate than the first, since it doesn't rely on trend data but instead actual convergence of iterative solving. Plus, that makes it way cooler.
To find the sagitta, you need to first get the radius, which is...
$R = L/(X_F*2)$
WHERE:
$R$ is the radius of the arc.
$L$ is the length of the arc.
$X_F$ is the final iteration of $X_{N+1}$ from the Newton-Raphson Method.
Finally, to find the sagitta from the radius and cord length...
If $L/c < pi/2$, then...
$h = R - sqrt(R^2 - (c/2)^2)$
If $L/c > pi/2$, then...
$h = R + sqrt(R^2 - (c/2)^2)$
WHERE:
$h$ is the sagitta or height of the arc.
$R$ is the arc radius.
$c$ is the cord length of the arc.
Using our example to solve the above equations yields...
$R = 20/(0.78668307*2) = 12.71159932$
Since $0.78668307 < pi/2$ ...
$h = 12.71159932 - sqrt(12.71159932^2 - (18/2)^2) = 3.73469800$
Using the old solver method first mentioned, the resulting sagitta was $3.78510859$ ... Let's solve for the error of the polynomial curve fit vs. Newton-Raphson method as a difference of new minus old over the cord length:
$(3.73469800 - 3.78510859)/18 = -0.28005883% $
So... the old method was not too far off, at least in comparison with my guessing equation. Glad I didn't steer too far off the course from perfect.
To make it all nice and neat, I created a spreadsheet and uploaded it to Google Docs. Here is the link:
https://drive.google.com/file/d/16x5tC17QS8Lj07CMIcRDtiMbe14vnX1d/view?usp=sharing
Download it, and open with Excel. Do not try to open on Google Docs, since I have used features specific to Excel. To use the spreadsheet, type in your arc length and cord length into cells B2 and B3, respectively. The Central Angle, Radius, and Sagitta will be solved in cells B8 (in radians, B9 for degrees), B10, and B11, respectively.
To make this new spreadsheet more hands-free, if the solver method comes up with $X_N=0$ within 16 decimals, the iterations will restart from initial guess, but multiply it by a factor of the number of iterations to the 2.5th power divided by 2.5 ... which, so you know, "2.5" was chosen because for some reason it works very nicely to reduce the number of iterations approaching zero vs. some other values I have tried.
2ND MAJOR UPDATE
Now that I have had some sleep, I realized that the sigmoidal curve I had used for guessing the half arc angle was indeed supposed to be assymetrical. In fact, it is near perfect of an assymetrical sigmoidal curve. It makes sense that my original symmetrical sigmoidal curve was going to have some larger error than necessary.
I was actually shocked to have approximated the answer so closely that it could arguably be used at a larger scale. Certainly, it could be used alone for what I need it for (estimating curve of a fabric). Here is where I have found the information on formulating an equation for an assymetrical sigmoidal curve:
https://www.graphpad.com/guides/prism/7/curve-fitting/index.htm?reg_asymmetric_dose_response_ec.htm
Here is the theoretical curve fit equation for obtaining half of the central arc angle from only the arc length and cord length:
$X = B + (T - B)/(1 + 10^{(E + log_{10}(2^{1/S} - 1)/H - ln(L/c - 1))*H})^S$
WHERE:
$X$ is the half of the central arc angle.
$B$ is the Bottom or lowest possible value of $X$.
$T$ is the Top or highest possible value of $X$.
$E$ is the logEC50 or about where 50% of the concentrations that give half-maximum effects... basically, it moves the curve's inflection point to the left if negative or right if positive.
$log_{10}()$ is a logarithmic function, base $10$.
$S$ is the Symmetry and is a number greater than zero that controls how pinched or loose the upper and lower curves are. Less than one indicates a more gradual curve towards $-infty$ than $+infty$.
$H$ is Hillslope - how steep the slope at the inflection point the curve is.
$ln()$ is a (natural) logarithmic function, base $e$.
$L$ is the arc length.
$c$ is the cord length.
Here are the values for the constants I have determined are most accurate, using Excel's solver tool:
$B = 0$
$T = pi$
$E = -0.559341467113653 cong -0.55934147$
$H = 0.410514378955745 cong 0.41051438$
$S = 0.524651055687417 cong 0.52465106$
This solves to the following equations:
$X = pi/(1 + 10^{(-0.55934147 + log_{10}(2^{1/0.52465106} - 1)/0.41051438 - ln(L/c - 1))*0.41051438})^{0.52465106}$
Math Code (ie., a Google Search):
2*pi/(1+10^((-0.55934147+log(2^(1/0.52465106)-1)/0.41051438-ln(L/c-1))*0.41051438))^0.52465106
Where L and c are replaced by real values.
Excel Code:
2*PI()/(1+10^((-0.55934147+LOG10(2^(1/0.52465106)-1)/0.41051438-LN(B2/B3-1))*0.41051438))^0.52465106
Where L and c are located in cells B2 and B3.
$R = L/(2*pi/(1 + 10^{(-0.55934147 + log_{10}(2^{1/0.52465106} - 1)/0.41051438 - ln(L/c - 1))*0.41051438})^{0.52465106})$
Math Code (ie., a Google Search):
L/(2*pi/(1+10^((-0.55934147+log(2^(1/0.52465106)-1)/0.41051438-ln(L/c-1))*0.41051438))^0.52465106)
Where L and c are replaced by real values.
Excel Code:
B2/(2*PI()/(1+10^((-0.55934147+LOG10(2^(1/0.52465106)-1)/0.41051438-LN(B2/B3-1))*0.41051438))^0.52465106)
Where L and c are located in cells B2 and B3.
If $L/c < pi/2$, Then...
$h = L/(2*pi/(1 + 10^{(-0.55934147 + log_{10}(2^{1/0.52465106} - 1)/0.41051438 - ln(L/c - 1))*0.41051438})^{0.52465106}) - sqrt((L/(2*pi/(1 + 10^{(-0.55934147 + log_{10}(2^{1/0.52465106} - 1)/0.41051438 - ln(L/c - 1))*0.41051438})^{0.52465106}))^2 - (c/2)^2)$
Math Code (ie., a Google Search):
L/(2*pi/(1+10^((-0.55934147+log(2^(1/0.52465106)-1)/0.41051438-ln(L/c-1))*0.41051438))^0.52465106)-sqrt((L/(2*pi/(1+10^((-0.55934147+log(2^(1/0.52465106)-1)/0.41051438-ln(L/c-1))*0.41051438))^0.52465106))^2-(c/2)^2)
Where L and c are replaced by real values.
Excel Code:
B2/(2*PI()/(1+10^((-0.55934147+LOG10(2^(1/0.52465106)-1)/0.41051438-LN(B2/B3-1))*0.41051438))^0.52465106)-SQRT((B2/(2*PI()/(1+10^((-0.55934147+LOG10(2^(1/0.52465106)-1)/0.41051438-LN(B2/B3-1))*0.41051438))^0.52465106))^2-(B3/2)^2)
Where L and c are located in cells B2 and B3.
If $L/c > pi/2$, Then...
$h = L/(2*pi/(1 + 10^{(-0.55934147 + log_{10}(2^{1/0.52465106} - 1)/0.41051438 + ln(L/c - 1))*0.41051438})^{0.52465106}) + sqrt((L/(2*pi/(1 + 10^{(-0.55934147 + log_{10}(2^{1/0.52465106} - 1)/0.41051438 - ln(L/c - 1))*0.41051438})^{0.52465106}))^2 - (c/2)^2)$
Math Code (ie., a Google Search):
L/(2*pi/(1+10^((-0.55934147+log(2^(1/0.52465106)-1)/0.41051438-ln(L/c-1))*0.41051438))^0.52465106)+sqrt((L/(2*pi/(1+10^((-0.55934147+log(2^(1/0.52465106)-1)/0.41051438-ln(L/c-1))*0.41051438))^0.52465106))^2-(c/2)^2)
Where L and c are replaced by real values.
Excel Code:
B2/(2*PI()/(1+10^((-0.55934147+LOG10(2^(1/0.52465106)-1)/0.41051438-LN(B2/B3-1))*0.41051438))^0.52465106)+SQRT((B2/(2*PI()/(1+10^((-0.55934147+LOG10(2^(1/0.52465106)-1)/0.41051438-LN(B2/B3-1))*0.41051438))^0.52465106))^2-(B3/2)^2)
Where L and c are located in cells B2 and B3.
This is quite accurate. For our example up above of $L=20$ and $c=18$, the error as a function of the difference between sagitta guessed vs. solved using the Newton-Raphson method divided by the cord length is only $0.0879% $. The $R^2$ value is a remarkable $0.99999834$ - much better than the first method as a whole. I believe I am the first to come up with this solution for this particular problem (at least publicly on here, since Stigler's law would have you believe different), so just call it the "Devin Approximation" since that name does not seem to be taken and is easier to say than the "Logarithmic Asymmetrical Sigmoidal Approximation".
Here is the Excel file from the first major update now incorporating the new guessing method and having an additional tab called the "Sagitta Solver - Devin Approx." just using the equations above:
https://drive.google.com/file/d/1u5cg56ziQY6_hX9FhWxqFHJjRDAFMGCq/view?usp=sharing
As with the previous files, they must be downloaded and opened in Excel, not to be opened in Google Spreadsheets, since the spreadsheet has features only available in Excel. To use it, enter in the values for arc length and cord length in cells B2 and B3, respectively, and that will solve for the central angle, radius, and sagitta in cells B16, B17, and B18, respectively on the first tab.
answered Dec 12 '18 at 21:47
Devin ThayerDevin Thayer
11
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Given the effort you put into writing this, it may be worth just a little more effort to put the formulas in MathJax format so people might actually read all of this. You can get started at math.stackexchange.com/help
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– David K
Jan 3 at 19:16
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Oh snap. Thank you. I will get right on it.
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– Devin Thayer
Jan 3 at 19:57
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All set :) fixed.
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– Devin Thayer
Jan 3 at 21:09
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Hint:
If the arc lenght is $r theta$ (where $theta$ is the angle subtending the arc $AB$) we have:
For the cord: $AB=2rsin (theta/2)$
For the sagitta: $s= r(1-cos(theta /2)$)
Can you prove these results using simple trigonometry on the circle of radius $r$?
If not see here.
answered Jun 5 '15 at 19:56
Emilio NovatiEmilio Novati
51.8k43474
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$begingroup$
The first one seems just like the definition of a Sine (Since AB/2 is one of the short sides of the triangle, and r-s is the other one). In the sagitta case, it's basically just the radius minus the length from center of the circle to the chord (r-s), so you get the sagitta. That's not exactly where my issue lies. I know AB, and I know h at any given time. So here's what I have : I know that $ R = frac{(c/2)^2 + h^2}{2h} $. My chord length is known (calling it c). So, my arc length will be $L = frac{(c/2)^2 + h^2}{2h} times arcsin{ frac{2hc}{h^2+c^2}}$. I can't solve this for h.
$endgroup$
– Luis Costa
Jun 5 '15 at 20:20
$begingroup$
sorry h = sagitta.
$endgroup$
– Luis Costa
Jun 5 '15 at 20:24
$begingroup$
maybe that I don't understand your question. But, if you know the cord $c$ and the sagitta $h$ you can find the radius $r$ solving the equation: $r^2=(c/2)^2+(r-h)^2$ than find the arc $theta$ using using the equation that gives the cord ( or the sagitta).
$endgroup$
– Emilio Novati
Jun 5 '15 at 20:39
add a comment |
$begingroup$
Hint:
If the arc lenght is $r theta$ (where $theta$ is the angle subtending the arc $AB$) we have:
For the cord: $AB=2rsin (theta/2)$
For the sagitta: $s= r(1-cos(theta /2)$)
Can you prove these results using simple trigonometry on the circle of radius $r$?
If not see here.
answered Jun 5 '15 at 19:56
Emilio NovatiEmilio Novati
51.8k43474
$endgroup$
$begingroup$
The first one seems just like the definition of a Sine (Since AB/2 is one of the short sides of the triangle, and r-s is the other one). In the sagitta case, it's basically just the radius minus the length from center of the circle to the chord (r-s), so you get the sagitta. That's not exactly where my issue lies. I know AB, and I know h at any given time. So here's what I have : I know that $ R = frac{(c/2)^2 + h^2}{2h} $. My chord length is known (calling it c). So, my arc length will be $L = frac{(c/2)^2 + h^2}{2h} times arcsin{ frac{2hc}{h^2+c^2}}$. I can't solve this for h.
$endgroup$
– Luis Costa
Jun 5 '15 at 20:20
$begingroup$
sorry h = sagitta.
$endgroup$
– Luis Costa
Jun 5 '15 at 20:24
$begingroup$
maybe that I don't understand your question. But, if you know the cord $c$ and the sagitta $h$ you can find the radius $r$ solving the equation: $r^2=(c/2)^2+(r-h)^2$ than find the arc $theta$ using using the equation that gives the cord ( or the sagitta).
$endgroup$
– Emilio Novati
Jun 5 '15 at 20:39
add a comment |
$begingroup$
Hint:
If the arc lenght is $r theta$ (where $theta$ is the angle subtending the arc $AB$) we have:
For the cord: $AB=2rsin (theta/2)$
For the sagitta: $s= r(1-cos(theta /2)$)
Can you prove these results using simple trigonometry on the circle of radius $r$?
If not see here.
answered Jun 5 '15 at 19:56
Emilio NovatiEmilio Novati
51.8k43474
$endgroup$
Hint:
If the arc lenght is $r theta$ (where $theta$ is the angle subtending the arc $AB$) we have:
For the cord: $AB=2rsin (theta/2)$
For the sagitta: $s= r(1-cos(theta /2)$)
Can you prove these results using simple trigonometry on the circle of radius $r$?
If not see here.
answered Jun 5 '15 at 19:56
Emilio NovatiEmilio Novati
51.8k43474
answered Jun 5 '15 at 19:56
Emilio NovatiEmilio Novati
51.8k43474
answered Jun 5 '15 at 19:56
Emilio NovatiEmilio Novati
51.8k43474
answered Jun 5 '15 at 19:56
Emilio NovatiEmilio Novati
51.8k43474
51.8k43474
$begingroup$
The first one seems just like the definition of a Sine (Since AB/2 is one of the short sides of the triangle, and r-s is the other one). In the sagitta case, it's basically just the radius minus the length from center of the circle to the chord (r-s), so you get the sagitta. That's not exactly where my issue lies. I know AB, and I know h at any given time. So here's what I have : I know that $ R = frac{(c/2)^2 + h^2}{2h} $. My chord length is known (calling it c). So, my arc length will be $L = frac{(c/2)^2 + h^2}{2h} times arcsin{ frac{2hc}{h^2+c^2}}$. I can't solve this for h.
$endgroup$
– Luis Costa
Jun 5 '15 at 20:20
$begingroup$
sorry h = sagitta.
$endgroup$
– Luis Costa
Jun 5 '15 at 20:24
$begingroup$
maybe that I don't understand your question. But, if you know the cord $c$ and the sagitta $h$ you can find the radius $r$ solving the equation: $r^2=(c/2)^2+(r-h)^2$ than find the arc $theta$ using using the equation that gives the cord ( or the sagitta).
$endgroup$
– Emilio Novati
Jun 5 '15 at 20:39
add a comment |
$begingroup$
The first one seems just like the definition of a Sine (Since AB/2 is one of the short sides of the triangle, and r-s is the other one). In the sagitta case, it's basically just the radius minus the length from center of the circle to the chord (r-s), so you get the sagitta. That's not exactly where my issue lies. I know AB, and I know h at any given time. So here's what I have : I know that $ R = frac{(c/2)^2 + h^2}{2h} $. My chord length is known (calling it c). So, my arc length will be $L = frac{(c/2)^2 + h^2}{2h} times arcsin{ frac{2hc}{h^2+c^2}}$. I can't solve this for h.
$endgroup$
– Luis Costa
Jun 5 '15 at 20:20
$begingroup$
sorry h = sagitta.
$endgroup$
– Luis Costa
Jun 5 '15 at 20:24
$begingroup$
maybe that I don't understand your question. But, if you know the cord $c$ and the sagitta $h$ you can find the radius $r$ solving the equation: $r^2=(c/2)^2+(r-h)^2$ than find the arc $theta$ using using the equation that gives the cord ( or the sagitta).
$endgroup$
– Emilio Novati
Jun 5 '15 at 20:39
$begingroup$
The first one seems just like the definition of a Sine (Since AB/2 is one of the short sides of the triangle, and r-s is the other one). In the sagitta case, it's basically just the radius minus the length from center of the circle to the chord (r-s), so you get the sagitta. That's not exactly where my issue lies. I know AB, and I know h at any given time. So here's what I have : I know that $ R = frac{(c/2)^2 + h^2}{2h} $. My chord length is known (calling it c). So, my arc length will be $L = frac{(c/2)^2 + h^2}{2h} times arcsin{ frac{2hc}{h^2+c^2}}$. I can't solve this for h.
$endgroup$
– Luis Costa
Jun 5 '15 at 20:20
$begingroup$
The first one seems just like the definition of a Sine (Since AB/2 is one of the short sides of the triangle, and r-s is the other one). In the sagitta case, it's basically just the radius minus the length from center of the circle to the chord (r-s), so you get the sagitta. That's not exactly where my issue lies. I know AB, and I know h at any given time. So here's what I have : I know that $ R = frac{(c/2)^2 + h^2}{2h} $. My chord length is known (calling it c). So, my arc length will be $L = frac{(c/2)^2 + h^2}{2h} times arcsin{ frac{2hc}{h^2+c^2}}$. I can't solve this for h.
$endgroup$
– Luis Costa
Jun 5 '15 at 20:20
$begingroup$
sorry h = sagitta.
$endgroup$
– Luis Costa
Jun 5 '15 at 20:24
$begingroup$
sorry h = sagitta.
$endgroup$
– Luis Costa
Jun 5 '15 at 20:24
$begingroup$
maybe that I don't understand your question. But, if you know the cord $c$ and the sagitta $h$ you can find the radius $r$ solving the equation: $r^2=(c/2)^2+(r-h)^2$ than find the arc $theta$ using using the equation that gives the cord ( or the sagitta).
$endgroup$
– Emilio Novati
Jun 5 '15 at 20:39
$begingroup$
maybe that I don't understand your question. But, if you know the cord $c$ and the sagitta $h$ you can find the radius $r$ solving the equation: $r^2=(c/2)^2+(r-h)^2$ than find the arc $theta$ using using the equation that gives the cord ( or the sagitta).
$endgroup$
– Emilio Novati
Jun 5 '15 at 20:39
add a comment |
$begingroup$
NOTE: SEE BELOW FOR A MAJOR UPDATE BEFORE YOU READ THIS FIRST PART.
Using several equation solver tools, I was unable to converge on an exact solution. Essentially, the equation $L=arcsin(frac{c}{h+frac{c^2}{4h}}*(h+frac{c^2}{4h}))$ has to be arranged to solve for $h$, which broke every solver tool I had (where $c$ is chord length, $h$ is height of segment or sagitta, and $L$ is arc length). Instead, I offer you a $R^2 = 0.9999$ solution from graphed data points.
I popped open Solidworks, a CAD program, which uses an iterative solver to determine this geometry, and I drew a circular segment, constraining at the chord length and arc length.
Image: Screenshot of segment showing arc length of 1.25 in Solidworks
Making sure the cord length was equal to $1$, I changed the arc length little by little and recorded the output height. This gave me the following table of values from a $L/c$ ratio of $1.01$ to $pi/2$:
x = L/c y = h/c
1.01 0.06132899
1.015 0.07516845
1.02 0.08686174
1.03 0.10654164
1.04 0.12320591
1.05 0.13795167
1.1 0.19652075
1.15 0.24241721
1.2 0.28189519
1.25 0.31735497
1.3 0.3500165
1.35 0.38059796
1.4 0.40956308
1.45 0.43723107
1.5 0.46383222
1.55 0.48953882
1.56 0.49458502
1.570796327 0.5
Here is the 5th order quadratic equation driven from that data using mycurvefit.com:
$y = -130.1039 + 489.8105*x - 737.3181*x^2 + 554.5809*x^3 - 208.0569*x^4 + ...$ $31.13006*x^5$
All you do is pop in the ratio $L/c$ for $x$ and then you get $y$, the ratio $h/c$. Since the chord length is known, and you have the ratio $h/c$, you can solve for the height by multiplying the ratio of $h/c$ by the chord length.
...but I kept going...
I wondered if someone reading this might have an application for the inverse segment. Things get less complicated as the ratio of $L/c$ approaches infinity.
Here is the data from $x = pi/2$ to $x = pi$:
x = L/c y = h/c
1.570796327 0.5
1.58 0.50458986
1.59 0.50955019
1.6 0.51448353
1.7 0.56248484
1.8 0.60845528
1.9 0.65281718
2 0.6958726
2.1 0.73784553
2.2 0.778907
2.3 0.81919066
2.4 0.85880306
2.5 0.89783052
2.6 0.93634401
2.7 0.97440257
2.8 1.01205584
2.9 1.04934598
3 1.08630908
3.14159265 1.13814656
This has a polynomial curve fit as follows:
$y = -0.5308871 + 0.8723522*x - 0.1662753*x^2 + 0.01838478*x^3$
Here is the data from $pi$ to $10$:
x = L/c y = h/c
3.14159265 1.13814656
3.25 1.17748049
3.5 1.26717904
3.75 1.35568642
4 1.44321347
4.5 1.61593815
5 1.78627783
5.5 1.9548322
6 2.12201017
7 2.45331872
8 2.78172909
9 3.10812811
10 3.43306996
Which has a polynomial curve fit as follows:
$y = 0.02093882 + 0.364244*x - 0.002337369*x^2$
The relationship between $L/c$ and h/c becomes very linear after this point, as $R^2$ values still are $0.9999$ or better just using a linear equation. Here is the data from $x = 10$ to $100$:
x = L/c y = h/c
10 3.43306996
15 5.04548886
20 6.64810187
25 8.24652077
50 16.21859143
75 24.1813035
100 32.14157177
The linear regression for this curve fit is as follows:
$y = 0.3188628*x + 0.2643024$
Beyond this point, I am not sure if the iterative solver is becoming unreliable due to rounding errors, so use the next information with a grain of salt as it approaches the capabilities of the software. Here is the data from $100$ to $10^{10}$:
x = L/c y = h/c
100 32.14157177
1000 318.627412
100000 31831.30692
10000000000 3183098862
The linear regression is as follows:
$y = 0.3183099*x + 0.3154708$
Despite the growing value of the intercept, it is relatively insignificant in light of the magnitude of the input or output and is probably just a growing rounding error as the software approaches its limitations in the iterative solver. Therefore, beyond $10^{10}$, you can simply use $m = pi^{-1}$ and $b = 0$ in the linear regression formula $y = m*x + b$:
$y = x/pi$
Or in other words, as $x$ approaches infinity, $m$ approaches the inverse of $pi$ and $b$ approaches zero. In practical terms, the smaller the chord is, the closer the sagitta becomes the diameter. I think that passes the smell test.
See this link to download an Excel file solver:
https://drive.google.com/file/d/1er5kIRyL6UahqqAAjlKTEVVhOeFyDZNd/view?usp=sharing
Download it if you want to use it. Don't bother editting or opening in Google Docs. I used features only available in Excel. Once open in Excel, just enter in the Arc and Chord lengths into cells B2 and B3, respectively, and out pops the Sigatta in cell B8.
MAJOR UPDATE
I have since found an additional way to solve this problem thanks to the following resources from 1728.org:
Circle Segment Calculator: http://www.1728.org/circsect.htm
Explanation of Newton's Method: http://www.1728.org/newton.htm
Explained on the second link, I was able to solve for the arc angle using the Newton-Raphson Method of Iterative solving. Usually, the solution converges within 8 decimal places within 4 iterations. However, if the guess is far too great or small, the method fails to find a valid solution (and breaks the solver for the first link). To prevent this, I created a method by which to estimate the initial guess for the half arc angle, mathematically.
$X_{Guess} = pi*(tanh((ln(L/c-1)+0.646543492637008)*0.373118408093598)+1)$
WHERE:
$X_{Guess}$ is an approximation for half the arc angle in radians.
$tanh()$ and $ln()$ are trigonometric and logarithmic functions, respectively.
$L$ is the arc length.
$c$ is the cord length.
The $R^2$ value is about $0.999005$, so this gives a pretty close guess to the arc angle. The largest error I have noticed on my chart is an error value of $2.815% $ as percentage of difference ($X_{Data}$ minus $X_{Guess}$) divided by $pi$. Because it is so close, you can use this as an input to solve for a much closer solution using the Newton-Raphson method as follows:
$f(X_N) = sin(X_N) - c*X_N/L = 0$
$f'(X_N) = cos(X_N) - c/L = 0$
$X_{N+1} = X_N - f(X)/f'(X)$
WHERE:
$f(X_N)$ is an equation of equality given the angle and the chord and arc lengths.
$sin()$ and $cos()$ are trigonometric functions.
$c$ is cord length.
$L$ is the arc length.
$f'(X_N)$ is the derivative of $f(X_N)$.
$X_{N+1}$ is the next iteration of X to solve using the Newton-Raphson Method.
Using the $X_{Guess}$ from the first equation, input it into those three equations to find the next $X_N$ and so on until the $X_N$'s stop changing. For instance, using $L=20$ and $c=18$...
$X_{Guess} = pi*(tanh((ln(20/18-1)+0.646543492637008)*0.373118408093598)+1) = 0.75140903$
$f(X_{Guess}) = sin(0.75140903) - 18*0.75140903/20 = 0.00640093$
$f'(X_{Guess}) = cos(0.75140903) - 18/20 = 0.75140903$
$X_1 = 0.75140903 - 0.00640093/0.75140903 = 0.74289047$
Inputting $X_1$ in place of $X_{Guess}$, $X_2$ in place of $X_1$ and so on...
$X_2 = 0.79071296$
$X_3 = 0.78671238$
$X_4 = 0.78668307$
$X_5 = 0.78668307$
And there we stop, since the solution has converged on 8 decimal places.
Error from guess as a difference ($X_5$ minus $X_{Guess}$) divided by $pi$:
$(0.75140903 - 0.78668307)/pi = -1.123% $
Which, with a 100 FT of material, that 1% could make a difference. Furthermore, this method is arguably more accurate than the first, since it doesn't rely on trend data but instead actual convergence of iterative solving. Plus, that makes it way cooler.
To find the sagitta, you need to first get the radius, which is...
$R = L/(X_F*2)$
WHERE:
$R$ is the radius of the arc.
$L$ is the length of the arc.
$X_F$ is the final iteration of $X_{N+1}$ from the Newton-Raphson Method.
Finally, to find the sagitta from the radius and cord length...
If $L/c < pi/2$, then...
$h = R - sqrt(R^2 - (c/2)^2)$
If $L/c > pi/2$, then...
$h = R + sqrt(R^2 - (c/2)^2)$
WHERE:
$h$ is the sagitta or height of the arc.
$R$ is the arc radius.
$c$ is the cord length of the arc.
Using our example to solve the above equations yields...
$R = 20/(0.78668307*2) = 12.71159932$
Since $0.78668307 < pi/2$ ...
$h = 12.71159932 - sqrt(12.71159932^2 - (18/2)^2) = 3.73469800$
Using the old solver method first mentioned, the resulting sagitta was $3.78510859$ ... Let's solve for the error of the polynomial curve fit vs. Newton-Raphson method as a difference of new minus old over the cord length:
$(3.73469800 - 3.78510859)/18 = -0.28005883% $
So... the old method was not too far off, at least in comparison with my guessing equation. Glad I didn't steer too far off the course from perfect.
To make it all nice and neat, I created a spreadsheet and uploaded it to Google Docs. Here is the link:
https://drive.google.com/file/d/16x5tC17QS8Lj07CMIcRDtiMbe14vnX1d/view?usp=sharing
Download it, and open with Excel. Do not try to open on Google Docs, since I have used features specific to Excel. To use the spreadsheet, type in your arc length and cord length into cells B2 and B3, respectively. The Central Angle, Radius, and Sagitta will be solved in cells B8 (in radians, B9 for degrees), B10, and B11, respectively.
To make this new spreadsheet more hands-free, if the solver method comes up with $X_N=0$ within 16 decimals, the iterations will restart from initial guess, but multiply it by a factor of the number of iterations to the 2.5th power divided by 2.5 ... which, so you know, "2.5" was chosen because for some reason it works very nicely to reduce the number of iterations approaching zero vs. some other values I have tried.
2ND MAJOR UPDATE
Now that I have had some sleep, I realized that the sigmoidal curve I had used for guessing the half arc angle was indeed supposed to be assymetrical. In fact, it is near perfect of an assymetrical sigmoidal curve. It makes sense that my original symmetrical sigmoidal curve was going to have some larger error than necessary.
I was actually shocked to have approximated the answer so closely that it could arguably be used at a larger scale. Certainly, it could be used alone for what I need it for (estimating curve of a fabric). Here is where I have found the information on formulating an equation for an assymetrical sigmoidal curve:
https://www.graphpad.com/guides/prism/7/curve-fitting/index.htm?reg_asymmetric_dose_response_ec.htm
Here is the theoretical curve fit equation for obtaining half of the central arc angle from only the arc length and cord length:
$X = B + (T - B)/(1 + 10^{(E + log_{10}(2^{1/S} - 1)/H - ln(L/c - 1))*H})^S$
WHERE:
$X$ is the half of the central arc angle.
$B$ is the Bottom or lowest possible value of $X$.
$T$ is the Top or highest possible value of $X$.
$E$ is the logEC50 or about where 50% of the concentrations that give half-maximum effects... basically, it moves the curve's inflection point to the left if negative or right if positive.
$log_{10}()$ is a logarithmic function, base $10$.
$S$ is the Symmetry and is a number greater than zero that controls how pinched or loose the upper and lower curves are. Less than one indicates a more gradual curve towards $-infty$ than $+infty$.
$H$ is Hillslope - how steep the slope at the inflection point the curve is.
$ln()$ is a (natural) logarithmic function, base $e$.
$L$ is the arc length.
$c$ is the cord length.
Here are the values for the constants I have determined are most accurate, using Excel's solver tool:
$B = 0$
$T = pi$
$E = -0.559341467113653 cong -0.55934147$
$H = 0.410514378955745 cong 0.41051438$
$S = 0.524651055687417 cong 0.52465106$
This solves to the following equations:
$X = pi/(1 + 10^{(-0.55934147 + log_{10}(2^{1/0.52465106} - 1)/0.41051438 - ln(L/c - 1))*0.41051438})^{0.52465106}$
Math Code (ie., a Google Search):
2*pi/(1+10^((-0.55934147+log(2^(1/0.52465106)-1)/0.41051438-ln(L/c-1))*0.41051438))^0.52465106
Where L and c are replaced by real values.
Excel Code:
2*PI()/(1+10^((-0.55934147+LOG10(2^(1/0.52465106)-1)/0.41051438-LN(B2/B3-1))*0.41051438))^0.52465106
Where L and c are located in cells B2 and B3.
$R = L/(2*pi/(1 + 10^{(-0.55934147 + log_{10}(2^{1/0.52465106} - 1)/0.41051438 - ln(L/c - 1))*0.41051438})^{0.52465106})$
Math Code (ie., a Google Search):
L/(2*pi/(1+10^((-0.55934147+log(2^(1/0.52465106)-1)/0.41051438-ln(L/c-1))*0.41051438))^0.52465106)
Where L and c are replaced by real values.
Excel Code:
B2/(2*PI()/(1+10^((-0.55934147+LOG10(2^(1/0.52465106)-1)/0.41051438-LN(B2/B3-1))*0.41051438))^0.52465106)
Where L and c are located in cells B2 and B3.
If $L/c < pi/2$, Then...
$h = L/(2*pi/(1 + 10^{(-0.55934147 + log_{10}(2^{1/0.52465106} - 1)/0.41051438 - ln(L/c - 1))*0.41051438})^{0.52465106}) - sqrt((L/(2*pi/(1 + 10^{(-0.55934147 + log_{10}(2^{1/0.52465106} - 1)/0.41051438 - ln(L/c - 1))*0.41051438})^{0.52465106}))^2 - (c/2)^2)$
Math Code (ie., a Google Search):
L/(2*pi/(1+10^((-0.55934147+log(2^(1/0.52465106)-1)/0.41051438-ln(L/c-1))*0.41051438))^0.52465106)-sqrt((L/(2*pi/(1+10^((-0.55934147+log(2^(1/0.52465106)-1)/0.41051438-ln(L/c-1))*0.41051438))^0.52465106))^2-(c/2)^2)
Where L and c are replaced by real values.
Excel Code:
B2/(2*PI()/(1+10^((-0.55934147+LOG10(2^(1/0.52465106)-1)/0.41051438-LN(B2/B3-1))*0.41051438))^0.52465106)-SQRT((B2/(2*PI()/(1+10^((-0.55934147+LOG10(2^(1/0.52465106)-1)/0.41051438-LN(B2/B3-1))*0.41051438))^0.52465106))^2-(B3/2)^2)
Where L and c are located in cells B2 and B3.
If $L/c > pi/2$, Then...
$h = L/(2*pi/(1 + 10^{(-0.55934147 + log_{10}(2^{1/0.52465106} - 1)/0.41051438 + ln(L/c - 1))*0.41051438})^{0.52465106}) + sqrt((L/(2*pi/(1 + 10^{(-0.55934147 + log_{10}(2^{1/0.52465106} - 1)/0.41051438 - ln(L/c - 1))*0.41051438})^{0.52465106}))^2 - (c/2)^2)$
Math Code (ie., a Google Search):
L/(2*pi/(1+10^((-0.55934147+log(2^(1/0.52465106)-1)/0.41051438-ln(L/c-1))*0.41051438))^0.52465106)+sqrt((L/(2*pi/(1+10^((-0.55934147+log(2^(1/0.52465106)-1)/0.41051438-ln(L/c-1))*0.41051438))^0.52465106))^2-(c/2)^2)
Where L and c are replaced by real values.
Excel Code:
B2/(2*PI()/(1+10^((-0.55934147+LOG10(2^(1/0.52465106)-1)/0.41051438-LN(B2/B3-1))*0.41051438))^0.52465106)+SQRT((B2/(2*PI()/(1+10^((-0.55934147+LOG10(2^(1/0.52465106)-1)/0.41051438-LN(B2/B3-1))*0.41051438))^0.52465106))^2-(B3/2)^2)
Where L and c are located in cells B2 and B3.
This is quite accurate. For our example up above of $L=20$ and $c=18$, the error as a function of the difference between sagitta guessed vs. solved using the Newton-Raphson method divided by the cord length is only $0.0879% $. The $R^2$ value is a remarkable $0.99999834$ - much better than the first method as a whole. I believe I am the first to come up with this solution for this particular problem (at least publicly on here, since Stigler's law would have you believe different), so just call it the "Devin Approximation" since that name does not seem to be taken and is easier to say than the "Logarithmic Asymmetrical Sigmoidal Approximation".
Here is the Excel file from the first major update now incorporating the new guessing method and having an additional tab called the "Sagitta Solver - Devin Approx." just using the equations above:
https://drive.google.com/file/d/1u5cg56ziQY6_hX9FhWxqFHJjRDAFMGCq/view?usp=sharing
As with the previous files, they must be downloaded and opened in Excel, not to be opened in Google Spreadsheets, since the spreadsheet has features only available in Excel. To use it, enter in the values for arc length and cord length in cells B2 and B3, respectively, and that will solve for the central angle, radius, and sagitta in cells B16, B17, and B18, respectively on the first tab.
answered Dec 12 '18 at 21:47
Devin ThayerDevin Thayer
11
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Given the effort you put into writing this, it may be worth just a little more effort to put the formulas in MathJax format so people might actually read all of this. You can get started at math.stackexchange.com/help
$endgroup$
– David K
Jan 3 at 19:16
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Oh snap. Thank you. I will get right on it.
$endgroup$
– Devin Thayer
Jan 3 at 19:57
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All set :) fixed.
$endgroup$
– Devin Thayer
Jan 3 at 21:09
add a comment |
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NOTE: SEE BELOW FOR A MAJOR UPDATE BEFORE YOU READ THIS FIRST PART.
Using several equation solver tools, I was unable to converge on an exact solution. Essentially, the equation $L=arcsin(frac{c}{h+frac{c^2}{4h}}*(h+frac{c^2}{4h}))$ has to be arranged to solve for $h$, which broke every solver tool I had (where $c$ is chord length, $h$ is height of segment or sagitta, and $L$ is arc length). Instead, I offer you a $R^2 = 0.9999$ solution from graphed data points.
I popped open Solidworks, a CAD program, which uses an iterative solver to determine this geometry, and I drew a circular segment, constraining at the chord length and arc length.
Image: Screenshot of segment showing arc length of 1.25 in Solidworks
Making sure the cord length was equal to $1$, I changed the arc length little by little and recorded the output height. This gave me the following table of values from a $L/c$ ratio of $1.01$ to $pi/2$:
x = L/c y = h/c
1.01 0.06132899
1.015 0.07516845
1.02 0.08686174
1.03 0.10654164
1.04 0.12320591
1.05 0.13795167
1.1 0.19652075
1.15 0.24241721
1.2 0.28189519
1.25 0.31735497
1.3 0.3500165
1.35 0.38059796
1.4 0.40956308
1.45 0.43723107
1.5 0.46383222
1.55 0.48953882
1.56 0.49458502
1.570796327 0.5
Here is the 5th order quadratic equation driven from that data using mycurvefit.com:
$y = -130.1039 + 489.8105*x - 737.3181*x^2 + 554.5809*x^3 - 208.0569*x^4 + ...$ $31.13006*x^5$
All you do is pop in the ratio $L/c$ for $x$ and then you get $y$, the ratio $h/c$. Since the chord length is known, and you have the ratio $h/c$, you can solve for the height by multiplying the ratio of $h/c$ by the chord length.
...but I kept going...
I wondered if someone reading this might have an application for the inverse segment. Things get less complicated as the ratio of $L/c$ approaches infinity.
Here is the data from $x = pi/2$ to $x = pi$:
x = L/c y = h/c
1.570796327 0.5
1.58 0.50458986
1.59 0.50955019
1.6 0.51448353
1.7 0.56248484
1.8 0.60845528
1.9 0.65281718
2 0.6958726
2.1 0.73784553
2.2 0.778907
2.3 0.81919066
2.4 0.85880306
2.5 0.89783052
2.6 0.93634401
2.7 0.97440257
2.8 1.01205584
2.9 1.04934598
3 1.08630908
3.14159265 1.13814656
This has a polynomial curve fit as follows:
$y = -0.5308871 + 0.8723522*x - 0.1662753*x^2 + 0.01838478*x^3$
Here is the data from $pi$ to $10$:
x = L/c y = h/c
3.14159265 1.13814656
3.25 1.17748049
3.5 1.26717904
3.75 1.35568642
4 1.44321347
4.5 1.61593815
5 1.78627783
5.5 1.9548322
6 2.12201017
7 2.45331872
8 2.78172909
9 3.10812811
10 3.43306996
Which has a polynomial curve fit as follows:
$y = 0.02093882 + 0.364244*x - 0.002337369*x^2$
The relationship between $L/c$ and h/c becomes very linear after this point, as $R^2$ values still are $0.9999$ or better just using a linear equation. Here is the data from $x = 10$ to $100$:
x = L/c y = h/c
10 3.43306996
15 5.04548886
20 6.64810187
25 8.24652077
50 16.21859143
75 24.1813035
100 32.14157177
The linear regression for this curve fit is as follows:
$y = 0.3188628*x + 0.2643024$
Beyond this point, I am not sure if the iterative solver is becoming unreliable due to rounding errors, so use the next information with a grain of salt as it approaches the capabilities of the software. Here is the data from $100$ to $10^{10}$:
x = L/c y = h/c
100 32.14157177
1000 318.627412
100000 31831.30692
10000000000 3183098862
The linear regression is as follows:
$y = 0.3183099*x + 0.3154708$
Despite the growing value of the intercept, it is relatively insignificant in light of the magnitude of the input or output and is probably just a growing rounding error as the software approaches its limitations in the iterative solver. Therefore, beyond $10^{10}$, you can simply use $m = pi^{-1}$ and $b = 0$ in the linear regression formula $y = m*x + b$:
$y = x/pi$
Or in other words, as $x$ approaches infinity, $m$ approaches the inverse of $pi$ and $b$ approaches zero. In practical terms, the smaller the chord is, the closer the sagitta becomes the diameter. I think that passes the smell test.
See this link to download an Excel file solver:
https://drive.google.com/file/d/1er5kIRyL6UahqqAAjlKTEVVhOeFyDZNd/view?usp=sharing
Download it if you want to use it. Don't bother editting or opening in Google Docs. I used features only available in Excel. Once open in Excel, just enter in the Arc and Chord lengths into cells B2 and B3, respectively, and out pops the Sigatta in cell B8.
MAJOR UPDATE
I have since found an additional way to solve this problem thanks to the following resources from 1728.org:
Circle Segment Calculator: http://www.1728.org/circsect.htm
Explanation of Newton's Method: http://www.1728.org/newton.htm
Explained on the second link, I was able to solve for the arc angle using the Newton-Raphson Method of Iterative solving. Usually, the solution converges within 8 decimal places within 4 iterations. However, if the guess is far too great or small, the method fails to find a valid solution (and breaks the solver for the first link). To prevent this, I created a method by which to estimate the initial guess for the half arc angle, mathematically.
$X_{Guess} = pi*(tanh((ln(L/c-1)+0.646543492637008)*0.373118408093598)+1)$
WHERE:
$X_{Guess}$ is an approximation for half the arc angle in radians.
$tanh()$ and $ln()$ are trigonometric and logarithmic functions, respectively.
$L$ is the arc length.
$c$ is the cord length.
The $R^2$ value is about $0.999005$, so this gives a pretty close guess to the arc angle. The largest error I have noticed on my chart is an error value of $2.815% $ as percentage of difference ($X_{Data}$ minus $X_{Guess}$) divided by $pi$. Because it is so close, you can use this as an input to solve for a much closer solution using the Newton-Raphson method as follows:
$f(X_N) = sin(X_N) - c*X_N/L = 0$
$f'(X_N) = cos(X_N) - c/L = 0$
$X_{N+1} = X_N - f(X)/f'(X)$
WHERE:
$f(X_N)$ is an equation of equality given the angle and the chord and arc lengths.
$sin()$ and $cos()$ are trigonometric functions.
$c$ is cord length.
$L$ is the arc length.
$f'(X_N)$ is the derivative of $f(X_N)$.
$X_{N+1}$ is the next iteration of X to solve using the Newton-Raphson Method.
Using the $X_{Guess}$ from the first equation, input it into those three equations to find the next $X_N$ and so on until the $X_N$'s stop changing. For instance, using $L=20$ and $c=18$...
$X_{Guess} = pi*(tanh((ln(20/18-1)+0.646543492637008)*0.373118408093598)+1) = 0.75140903$
$f(X_{Guess}) = sin(0.75140903) - 18*0.75140903/20 = 0.00640093$
$f'(X_{Guess}) = cos(0.75140903) - 18/20 = 0.75140903$
$X_1 = 0.75140903 - 0.00640093/0.75140903 = 0.74289047$
Inputting $X_1$ in place of $X_{Guess}$, $X_2$ in place of $X_1$ and so on...
$X_2 = 0.79071296$
$X_3 = 0.78671238$
$X_4 = 0.78668307$
$X_5 = 0.78668307$
And there we stop, since the solution has converged on 8 decimal places.
Error from guess as a difference ($X_5$ minus $X_{Guess}$) divided by $pi$:
$(0.75140903 - 0.78668307)/pi = -1.123% $
Which, with a 100 FT of material, that 1% could make a difference. Furthermore, this method is arguably more accurate than the first, since it doesn't rely on trend data but instead actual convergence of iterative solving. Plus, that makes it way cooler.
To find the sagitta, you need to first get the radius, which is...
$R = L/(X_F*2)$
WHERE:
$R$ is the radius of the arc.
$L$ is the length of the arc.
$X_F$ is the final iteration of $X_{N+1}$ from the Newton-Raphson Method.
Finally, to find the sagitta from the radius and cord length...
If $L/c < pi/2$, then...
$h = R - sqrt(R^2 - (c/2)^2)$
If $L/c > pi/2$, then...
$h = R + sqrt(R^2 - (c/2)^2)$
WHERE:
$h$ is the sagitta or height of the arc.
$R$ is the arc radius.
$c$ is the cord length of the arc.
Using our example to solve the above equations yields...
$R = 20/(0.78668307*2) = 12.71159932$
Since $0.78668307 < pi/2$ ...
$h = 12.71159932 - sqrt(12.71159932^2 - (18/2)^2) = 3.73469800$
Using the old solver method first mentioned, the resulting sagitta was $3.78510859$ ... Let's solve for the error of the polynomial curve fit vs. Newton-Raphson method as a difference of new minus old over the cord length:
$(3.73469800 - 3.78510859)/18 = -0.28005883% $
So... the old method was not too far off, at least in comparison with my guessing equation. Glad I didn't steer too far off the course from perfect.
To make it all nice and neat, I created a spreadsheet and uploaded it to Google Docs. Here is the link:
https://drive.google.com/file/d/16x5tC17QS8Lj07CMIcRDtiMbe14vnX1d/view?usp=sharing
Download it, and open with Excel. Do not try to open on Google Docs, since I have used features specific to Excel. To use the spreadsheet, type in your arc length and cord length into cells B2 and B3, respectively. The Central Angle, Radius, and Sagitta will be solved in cells B8 (in radians, B9 for degrees), B10, and B11, respectively.
To make this new spreadsheet more hands-free, if the solver method comes up with $X_N=0$ within 16 decimals, the iterations will restart from initial guess, but multiply it by a factor of the number of iterations to the 2.5th power divided by 2.5 ... which, so you know, "2.5" was chosen because for some reason it works very nicely to reduce the number of iterations approaching zero vs. some other values I have tried.
2ND MAJOR UPDATE
Now that I have had some sleep, I realized that the sigmoidal curve I had used for guessing the half arc angle was indeed supposed to be assymetrical. In fact, it is near perfect of an assymetrical sigmoidal curve. It makes sense that my original symmetrical sigmoidal curve was going to have some larger error than necessary.
I was actually shocked to have approximated the answer so closely that it could arguably be used at a larger scale. Certainly, it could be used alone for what I need it for (estimating curve of a fabric). Here is where I have found the information on formulating an equation for an assymetrical sigmoidal curve:
https://www.graphpad.com/guides/prism/7/curve-fitting/index.htm?reg_asymmetric_dose_response_ec.htm
Here is the theoretical curve fit equation for obtaining half of the central arc angle from only the arc length and cord length:
$X = B + (T - B)/(1 + 10^{(E + log_{10}(2^{1/S} - 1)/H - ln(L/c - 1))*H})^S$
WHERE:
$X$ is the half of the central arc angle.
$B$ is the Bottom or lowest possible value of $X$.
$T$ is the Top or highest possible value of $X$.
$E$ is the logEC50 or about where 50% of the concentrations that give half-maximum effects... basically, it moves the curve's inflection point to the left if negative or right if positive.
$log_{10}()$ is a logarithmic function, base $10$.
$S$ is the Symmetry and is a number greater than zero that controls how pinched or loose the upper and lower curves are. Less than one indicates a more gradual curve towards $-infty$ than $+infty$.
$H$ is Hillslope - how steep the slope at the inflection point the curve is.
$ln()$ is a (natural) logarithmic function, base $e$.
$L$ is the arc length.
$c$ is the cord length.
Here are the values for the constants I have determined are most accurate, using Excel's solver tool:
$B = 0$
$T = pi$
$E = -0.559341467113653 cong -0.55934147$
$H = 0.410514378955745 cong 0.41051438$
$S = 0.524651055687417 cong 0.52465106$
This solves to the following equations:
$X = pi/(1 + 10^{(-0.55934147 + log_{10}(2^{1/0.52465106} - 1)/0.41051438 - ln(L/c - 1))*0.41051438})^{0.52465106}$
Math Code (ie., a Google Search):
2*pi/(1+10^((-0.55934147+log(2^(1/0.52465106)-1)/0.41051438-ln(L/c-1))*0.41051438))^0.52465106
Where L and c are replaced by real values.
Excel Code:
2*PI()/(1+10^((-0.55934147+LOG10(2^(1/0.52465106)-1)/0.41051438-LN(B2/B3-1))*0.41051438))^0.52465106
Where L and c are located in cells B2 and B3.
$R = L/(2*pi/(1 + 10^{(-0.55934147 + log_{10}(2^{1/0.52465106} - 1)/0.41051438 - ln(L/c - 1))*0.41051438})^{0.52465106})$
Math Code (ie., a Google Search):
L/(2*pi/(1+10^((-0.55934147+log(2^(1/0.52465106)-1)/0.41051438-ln(L/c-1))*0.41051438))^0.52465106)
Where L and c are replaced by real values.
Excel Code:
B2/(2*PI()/(1+10^((-0.55934147+LOG10(2^(1/0.52465106)-1)/0.41051438-LN(B2/B3-1))*0.41051438))^0.52465106)
Where L and c are located in cells B2 and B3.
If $L/c < pi/2$, Then...
$h = L/(2*pi/(1 + 10^{(-0.55934147 + log_{10}(2^{1/0.52465106} - 1)/0.41051438 - ln(L/c - 1))*0.41051438})^{0.52465106}) - sqrt((L/(2*pi/(1 + 10^{(-0.55934147 + log_{10}(2^{1/0.52465106} - 1)/0.41051438 - ln(L/c - 1))*0.41051438})^{0.52465106}))^2 - (c/2)^2)$
Math Code (ie., a Google Search):
L/(2*pi/(1+10^((-0.55934147+log(2^(1/0.52465106)-1)/0.41051438-ln(L/c-1))*0.41051438))^0.52465106)-sqrt((L/(2*pi/(1+10^((-0.55934147+log(2^(1/0.52465106)-1)/0.41051438-ln(L/c-1))*0.41051438))^0.52465106))^2-(c/2)^2)
Where L and c are replaced by real values.
Excel Code:
B2/(2*PI()/(1+10^((-0.55934147+LOG10(2^(1/0.52465106)-1)/0.41051438-LN(B2/B3-1))*0.41051438))^0.52465106)-SQRT((B2/(2*PI()/(1+10^((-0.55934147+LOG10(2^(1/0.52465106)-1)/0.41051438-LN(B2/B3-1))*0.41051438))^0.52465106))^2-(B3/2)^2)
Where L and c are located in cells B2 and B3.
If $L/c > pi/2$, Then...
$h = L/(2*pi/(1 + 10^{(-0.55934147 + log_{10}(2^{1/0.52465106} - 1)/0.41051438 + ln(L/c - 1))*0.41051438})^{0.52465106}) + sqrt((L/(2*pi/(1 + 10^{(-0.55934147 + log_{10}(2^{1/0.52465106} - 1)/0.41051438 - ln(L/c - 1))*0.41051438})^{0.52465106}))^2 - (c/2)^2)$
Math Code (ie., a Google Search):
L/(2*pi/(1+10^((-0.55934147+log(2^(1/0.52465106)-1)/0.41051438-ln(L/c-1))*0.41051438))^0.52465106)+sqrt((L/(2*pi/(1+10^((-0.55934147+log(2^(1/0.52465106)-1)/0.41051438-ln(L/c-1))*0.41051438))^0.52465106))^2-(c/2)^2)
Where L and c are replaced by real values.
Excel Code:
B2/(2*PI()/(1+10^((-0.55934147+LOG10(2^(1/0.52465106)-1)/0.41051438-LN(B2/B3-1))*0.41051438))^0.52465106)+SQRT((B2/(2*PI()/(1+10^((-0.55934147+LOG10(2^(1/0.52465106)-1)/0.41051438-LN(B2/B3-1))*0.41051438))^0.52465106))^2-(B3/2)^2)
Where L and c are located in cells B2 and B3.
This is quite accurate. For our example up above of $L=20$ and $c=18$, the error as a function of the difference between sagitta guessed vs. solved using the Newton-Raphson method divided by the cord length is only $0.0879% $. The $R^2$ value is a remarkable $0.99999834$ - much better than the first method as a whole. I believe I am the first to come up with this solution for this particular problem (at least publicly on here, since Stigler's law would have you believe different), so just call it the "Devin Approximation" since that name does not seem to be taken and is easier to say than the "Logarithmic Asymmetrical Sigmoidal Approximation".
Here is the Excel file from the first major update now incorporating the new guessing method and having an additional tab called the "Sagitta Solver - Devin Approx." just using the equations above:
https://drive.google.com/file/d/1u5cg56ziQY6_hX9FhWxqFHJjRDAFMGCq/view?usp=sharing
As with the previous files, they must be downloaded and opened in Excel, not to be opened in Google Spreadsheets, since the spreadsheet has features only available in Excel. To use it, enter in the values for arc length and cord length in cells B2 and B3, respectively, and that will solve for the central angle, radius, and sagitta in cells B16, B17, and B18, respectively on the first tab.
answered Dec 12 '18 at 21:47
Devin ThayerDevin Thayer
11
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Given the effort you put into writing this, it may be worth just a little more effort to put the formulas in MathJax format so people might actually read all of this. You can get started at math.stackexchange.com/help
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– David K
Jan 3 at 19:16
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Oh snap. Thank you. I will get right on it.
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– Devin Thayer
Jan 3 at 19:57
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All set :) fixed.
$endgroup$
– Devin Thayer
Jan 3 at 21:09
add a comment |
$begingroup$
NOTE: SEE BELOW FOR A MAJOR UPDATE BEFORE YOU READ THIS FIRST PART.
Using several equation solver tools, I was unable to converge on an exact solution. Essentially, the equation $L=arcsin(frac{c}{h+frac{c^2}{4h}}*(h+frac{c^2}{4h}))$ has to be arranged to solve for $h$, which broke every solver tool I had (where $c$ is chord length, $h$ is height of segment or sagitta, and $L$ is arc length). Instead, I offer you a $R^2 = 0.9999$ solution from graphed data points.
I popped open Solidworks, a CAD program, which uses an iterative solver to determine this geometry, and I drew a circular segment, constraining at the chord length and arc length.
Image: Screenshot of segment showing arc length of 1.25 in Solidworks
Making sure the cord length was equal to $1$, I changed the arc length little by little and recorded the output height. This gave me the following table of values from a $L/c$ ratio of $1.01$ to $pi/2$:
x = L/c y = h/c
1.01 0.06132899
1.015 0.07516845
1.02 0.08686174
1.03 0.10654164
1.04 0.12320591
1.05 0.13795167
1.1 0.19652075
1.15 0.24241721
1.2 0.28189519
1.25 0.31735497
1.3 0.3500165
1.35 0.38059796
1.4 0.40956308
1.45 0.43723107
1.5 0.46383222
1.55 0.48953882
1.56 0.49458502
1.570796327 0.5
Here is the 5th order quadratic equation driven from that data using mycurvefit.com:
$y = -130.1039 + 489.8105*x - 737.3181*x^2 + 554.5809*x^3 - 208.0569*x^4 + ...$ $31.13006*x^5$
All you do is pop in the ratio $L/c$ for $x$ and then you get $y$, the ratio $h/c$. Since the chord length is known, and you have the ratio $h/c$, you can solve for the height by multiplying the ratio of $h/c$ by the chord length.
...but I kept going...
I wondered if someone reading this might have an application for the inverse segment. Things get less complicated as the ratio of $L/c$ approaches infinity.
Here is the data from $x = pi/2$ to $x = pi$:
x = L/c y = h/c
1.570796327 0.5
1.58 0.50458986
1.59 0.50955019
1.6 0.51448353
1.7 0.56248484
1.8 0.60845528
1.9 0.65281718
2 0.6958726
2.1 0.73784553
2.2 0.778907
2.3 0.81919066
2.4 0.85880306
2.5 0.89783052
2.6 0.93634401
2.7 0.97440257
2.8 1.01205584
2.9 1.04934598
3 1.08630908
3.14159265 1.13814656
This has a polynomial curve fit as follows:
$y = -0.5308871 + 0.8723522*x - 0.1662753*x^2 + 0.01838478*x^3$
Here is the data from $pi$ to $10$:
x = L/c y = h/c
3.14159265 1.13814656
3.25 1.17748049
3.5 1.26717904
3.75 1.35568642
4 1.44321347
4.5 1.61593815
5 1.78627783
5.5 1.9548322
6 2.12201017
7 2.45331872
8 2.78172909
9 3.10812811
10 3.43306996
Which has a polynomial curve fit as follows:
$y = 0.02093882 + 0.364244*x - 0.002337369*x^2$
The relationship between $L/c$ and h/c becomes very linear after this point, as $R^2$ values still are $0.9999$ or better just using a linear equation. Here is the data from $x = 10$ to $100$:
x = L/c y = h/c
10 3.43306996
15 5.04548886
20 6.64810187
25 8.24652077
50 16.21859143
75 24.1813035
100 32.14157177
The linear regression for this curve fit is as follows:
$y = 0.3188628*x + 0.2643024$
Beyond this point, I am not sure if the iterative solver is becoming unreliable due to rounding errors, so use the next information with a grain of salt as it approaches the capabilities of the software. Here is the data from $100$ to $10^{10}$:
x = L/c y = h/c
100 32.14157177
1000 318.627412
100000 31831.30692
10000000000 3183098862
The linear regression is as follows:
$y = 0.3183099*x + 0.3154708$
Despite the growing value of the intercept, it is relatively insignificant in light of the magnitude of the input or output and is probably just a growing rounding error as the software approaches its limitations in the iterative solver. Therefore, beyond $10^{10}$, you can simply use $m = pi^{-1}$ and $b = 0$ in the linear regression formula $y = m*x + b$:
$y = x/pi$
Or in other words, as $x$ approaches infinity, $m$ approaches the inverse of $pi$ and $b$ approaches zero. In practical terms, the smaller the chord is, the closer the sagitta becomes the diameter. I think that passes the smell test.
See this link to download an Excel file solver:
https://drive.google.com/file/d/1er5kIRyL6UahqqAAjlKTEVVhOeFyDZNd/view?usp=sharing
Download it if you want to use it. Don't bother editting or opening in Google Docs. I used features only available in Excel. Once open in Excel, just enter in the Arc and Chord lengths into cells B2 and B3, respectively, and out pops the Sigatta in cell B8.
MAJOR UPDATE
I have since found an additional way to solve this problem thanks to the following resources from 1728.org:
Circle Segment Calculator: http://www.1728.org/circsect.htm
Explanation of Newton's Method: http://www.1728.org/newton.htm
Explained on the second link, I was able to solve for the arc angle using the Newton-Raphson Method of Iterative solving. Usually, the solution converges within 8 decimal places within 4 iterations. However, if the guess is far too great or small, the method fails to find a valid solution (and breaks the solver for the first link). To prevent this, I created a method by which to estimate the initial guess for the half arc angle, mathematically.
$X_{Guess} = pi*(tanh((ln(L/c-1)+0.646543492637008)*0.373118408093598)+1)$
WHERE:
$X_{Guess}$ is an approximation for half the arc angle in radians.
$tanh()$ and $ln()$ are trigonometric and logarithmic functions, respectively.
$L$ is the arc length.
$c$ is the cord length.
The $R^2$ value is about $0.999005$, so this gives a pretty close guess to the arc angle. The largest error I have noticed on my chart is an error value of $2.815% $ as percentage of difference ($X_{Data}$ minus $X_{Guess}$) divided by $pi$. Because it is so close, you can use this as an input to solve for a much closer solution using the Newton-Raphson method as follows:
$f(X_N) = sin(X_N) - c*X_N/L = 0$
$f'(X_N) = cos(X_N) - c/L = 0$
$X_{N+1} = X_N - f(X)/f'(X)$
WHERE:
$f(X_N)$ is an equation of equality given the angle and the chord and arc lengths.
$sin()$ and $cos()$ are trigonometric functions.
$c$ is cord length.
$L$ is the arc length.
$f'(X_N)$ is the derivative of $f(X_N)$.
$X_{N+1}$ is the next iteration of X to solve using the Newton-Raphson Method.
Using the $X_{Guess}$ from the first equation, input it into those three equations to find the next $X_N$ and so on until the $X_N$'s stop changing. For instance, using $L=20$ and $c=18$...
$X_{Guess} = pi*(tanh((ln(20/18-1)+0.646543492637008)*0.373118408093598)+1) = 0.75140903$
$f(X_{Guess}) = sin(0.75140903) - 18*0.75140903/20 = 0.00640093$
$f'(X_{Guess}) = cos(0.75140903) - 18/20 = 0.75140903$
$X_1 = 0.75140903 - 0.00640093/0.75140903 = 0.74289047$
Inputting $X_1$ in place of $X_{Guess}$, $X_2$ in place of $X_1$ and so on...
$X_2 = 0.79071296$
$X_3 = 0.78671238$
$X_4 = 0.78668307$
$X_5 = 0.78668307$
And there we stop, since the solution has converged on 8 decimal places.
Error from guess as a difference ($X_5$ minus $X_{Guess}$) divided by $pi$:
$(0.75140903 - 0.78668307)/pi = -1.123% $
Which, with a 100 FT of material, that 1% could make a difference. Furthermore, this method is arguably more accurate than the first, since it doesn't rely on trend data but instead actual convergence of iterative solving. Plus, that makes it way cooler.
To find the sagitta, you need to first get the radius, which is...
$R = L/(X_F*2)$
WHERE:
$R$ is the radius of the arc.
$L$ is the length of the arc.
$X_F$ is the final iteration of $X_{N+1}$ from the Newton-Raphson Method.
Finally, to find the sagitta from the radius and cord length...
If $L/c < pi/2$, then...
$h = R - sqrt(R^2 - (c/2)^2)$
If $L/c > pi/2$, then...
$h = R + sqrt(R^2 - (c/2)^2)$
WHERE:
$h$ is the sagitta or height of the arc.
$R$ is the arc radius.
$c$ is the cord length of the arc.
Using our example to solve the above equations yields...
$R = 20/(0.78668307*2) = 12.71159932$
Since $0.78668307 < pi/2$ ...
$h = 12.71159932 - sqrt(12.71159932^2 - (18/2)^2) = 3.73469800$
Using the old solver method first mentioned, the resulting sagitta was $3.78510859$ ... Let's solve for the error of the polynomial curve fit vs. Newton-Raphson method as a difference of new minus old over the cord length:
$(3.73469800 - 3.78510859)/18 = -0.28005883% $
So... the old method was not too far off, at least in comparison with my guessing equation. Glad I didn't steer too far off the course from perfect.
To make it all nice and neat, I created a spreadsheet and uploaded it to Google Docs. Here is the link:
https://drive.google.com/file/d/16x5tC17QS8Lj07CMIcRDtiMbe14vnX1d/view?usp=sharing
Download it, and open with Excel. Do not try to open on Google Docs, since I have used features specific to Excel. To use the spreadsheet, type in your arc length and cord length into cells B2 and B3, respectively. The Central Angle, Radius, and Sagitta will be solved in cells B8 (in radians, B9 for degrees), B10, and B11, respectively.
To make this new spreadsheet more hands-free, if the solver method comes up with $X_N=0$ within 16 decimals, the iterations will restart from initial guess, but multiply it by a factor of the number of iterations to the 2.5th power divided by 2.5 ... which, so you know, "2.5" was chosen because for some reason it works very nicely to reduce the number of iterations approaching zero vs. some other values I have tried.
2ND MAJOR UPDATE
Now that I have had some sleep, I realized that the sigmoidal curve I had used for guessing the half arc angle was indeed supposed to be assymetrical. In fact, it is near perfect of an assymetrical sigmoidal curve. It makes sense that my original symmetrical sigmoidal curve was going to have some larger error than necessary.
I was actually shocked to have approximated the answer so closely that it could arguably be used at a larger scale. Certainly, it could be used alone for what I need it for (estimating curve of a fabric). Here is where I have found the information on formulating an equation for an assymetrical sigmoidal curve:
https://www.graphpad.com/guides/prism/7/curve-fitting/index.htm?reg_asymmetric_dose_response_ec.htm
Here is the theoretical curve fit equation for obtaining half of the central arc angle from only the arc length and cord length:
$X = B + (T - B)/(1 + 10^{(E + log_{10}(2^{1/S} - 1)/H - ln(L/c - 1))*H})^S$
WHERE:
$X$ is the half of the central arc angle.
$B$ is the Bottom or lowest possible value of $X$.
$T$ is the Top or highest possible value of $X$.
$E$ is the logEC50 or about where 50% of the concentrations that give half-maximum effects... basically, it moves the curve's inflection point to the left if negative or right if positive.
$log_{10}()$ is a logarithmic function, base $10$.
$S$ is the Symmetry and is a number greater than zero that controls how pinched or loose the upper and lower curves are. Less than one indicates a more gradual curve towards $-infty$ than $+infty$.
$H$ is Hillslope - how steep the slope at the inflection point the curve is.
$ln()$ is a (natural) logarithmic function, base $e$.
$L$ is the arc length.
$c$ is the cord length.
Here are the values for the constants I have determined are most accurate, using Excel's solver tool:
$B = 0$
$T = pi$
$E = -0.559341467113653 cong -0.55934147$
$H = 0.410514378955745 cong 0.41051438$
$S = 0.524651055687417 cong 0.52465106$
This solves to the following equations:
$X = pi/(1 + 10^{(-0.55934147 + log_{10}(2^{1/0.52465106} - 1)/0.41051438 - ln(L/c - 1))*0.41051438})^{0.52465106}$
Math Code (ie., a Google Search):
2*pi/(1+10^((-0.55934147+log(2^(1/0.52465106)-1)/0.41051438-ln(L/c-1))*0.41051438))^0.52465106
Where L and c are replaced by real values.
Excel Code:
2*PI()/(1+10^((-0.55934147+LOG10(2^(1/0.52465106)-1)/0.41051438-LN(B2/B3-1))*0.41051438))^0.52465106
Where L and c are located in cells B2 and B3.
$R = L/(2*pi/(1 + 10^{(-0.55934147 + log_{10}(2^{1/0.52465106} - 1)/0.41051438 - ln(L/c - 1))*0.41051438})^{0.52465106})$
Math Code (ie., a Google Search):
L/(2*pi/(1+10^((-0.55934147+log(2^(1/0.52465106)-1)/0.41051438-ln(L/c-1))*0.41051438))^0.52465106)
Where L and c are replaced by real values.
Excel Code:
B2/(2*PI()/(1+10^((-0.55934147+LOG10(2^(1/0.52465106)-1)/0.41051438-LN(B2/B3-1))*0.41051438))^0.52465106)
Where L and c are located in cells B2 and B3.
If $L/c < pi/2$, Then...
$h = L/(2*pi/(1 + 10^{(-0.55934147 + log_{10}(2^{1/0.52465106} - 1)/0.41051438 - ln(L/c - 1))*0.41051438})^{0.52465106}) - sqrt((L/(2*pi/(1 + 10^{(-0.55934147 + log_{10}(2^{1/0.52465106} - 1)/0.41051438 - ln(L/c - 1))*0.41051438})^{0.52465106}))^2 - (c/2)^2)$
Math Code (ie., a Google Search):
L/(2*pi/(1+10^((-0.55934147+log(2^(1/0.52465106)-1)/0.41051438-ln(L/c-1))*0.41051438))^0.52465106)-sqrt((L/(2*pi/(1+10^((-0.55934147+log(2^(1/0.52465106)-1)/0.41051438-ln(L/c-1))*0.41051438))^0.52465106))^2-(c/2)^2)
Where L and c are replaced by real values.
Excel Code:
B2/(2*PI()/(1+10^((-0.55934147+LOG10(2^(1/0.52465106)-1)/0.41051438-LN(B2/B3-1))*0.41051438))^0.52465106)-SQRT((B2/(2*PI()/(1+10^((-0.55934147+LOG10(2^(1/0.52465106)-1)/0.41051438-LN(B2/B3-1))*0.41051438))^0.52465106))^2-(B3/2)^2)
Where L and c are located in cells B2 and B3.
If $L/c > pi/2$, Then...
$h = L/(2*pi/(1 + 10^{(-0.55934147 + log_{10}(2^{1/0.52465106} - 1)/0.41051438 + ln(L/c - 1))*0.41051438})^{0.52465106}) + sqrt((L/(2*pi/(1 + 10^{(-0.55934147 + log_{10}(2^{1/0.52465106} - 1)/0.41051438 - ln(L/c - 1))*0.41051438})^{0.52465106}))^2 - (c/2)^2)$
Math Code (ie., a Google Search):
L/(2*pi/(1+10^((-0.55934147+log(2^(1/0.52465106)-1)/0.41051438-ln(L/c-1))*0.41051438))^0.52465106)+sqrt((L/(2*pi/(1+10^((-0.55934147+log(2^(1/0.52465106)-1)/0.41051438-ln(L/c-1))*0.41051438))^0.52465106))^2-(c/2)^2)
Where L and c are replaced by real values.
Excel Code:
B2/(2*PI()/(1+10^((-0.55934147+LOG10(2^(1/0.52465106)-1)/0.41051438-LN(B2/B3-1))*0.41051438))^0.52465106)+SQRT((B2/(2*PI()/(1+10^((-0.55934147+LOG10(2^(1/0.52465106)-1)/0.41051438-LN(B2/B3-1))*0.41051438))^0.52465106))^2-(B3/2)^2)
Where L and c are located in cells B2 and B3.
This is quite accurate. For our example up above of $L=20$ and $c=18$, the error as a function of the difference between sagitta guessed vs. solved using the Newton-Raphson method divided by the cord length is only $0.0879% $. The $R^2$ value is a remarkable $0.99999834$ - much better than the first method as a whole. I believe I am the first to come up with this solution for this particular problem (at least publicly on here, since Stigler's law would have you believe different), so just call it the "Devin Approximation" since that name does not seem to be taken and is easier to say than the "Logarithmic Asymmetrical Sigmoidal Approximation".
Here is the Excel file from the first major update now incorporating the new guessing method and having an additional tab called the "Sagitta Solver - Devin Approx." just using the equations above:
https://drive.google.com/file/d/1u5cg56ziQY6_hX9FhWxqFHJjRDAFMGCq/view?usp=sharing
As with the previous files, they must be downloaded and opened in Excel, not to be opened in Google Spreadsheets, since the spreadsheet has features only available in Excel. To use it, enter in the values for arc length and cord length in cells B2 and B3, respectively, and that will solve for the central angle, radius, and sagitta in cells B16, B17, and B18, respectively on the first tab.
answered Dec 12 '18 at 21:47
Devin ThayerDevin Thayer
11
$endgroup$
NOTE: SEE BELOW FOR A MAJOR UPDATE BEFORE YOU READ THIS FIRST PART.
Using several equation solver tools, I was unable to converge on an exact solution. Essentially, the equation $L=arcsin(frac{c}{h+frac{c^2}{4h}}*(h+frac{c^2}{4h}))$ has to be arranged to solve for $h$, which broke every solver tool I had (where $c$ is chord length, $h$ is height of segment or sagitta, and $L$ is arc length). Instead, I offer you a $R^2 = 0.9999$ solution from graphed data points.
I popped open Solidworks, a CAD program, which uses an iterative solver to determine this geometry, and I drew a circular segment, constraining at the chord length and arc length.
Image: Screenshot of segment showing arc length of 1.25 in Solidworks
Making sure the cord length was equal to $1$, I changed the arc length little by little and recorded the output height. This gave me the following table of values from a $L/c$ ratio of $1.01$ to $pi/2$:
x = L/c y = h/c
1.01 0.06132899
1.015 0.07516845
1.02 0.08686174
1.03 0.10654164
1.04 0.12320591
1.05 0.13795167
1.1 0.19652075
1.15 0.24241721
1.2 0.28189519
1.25 0.31735497
1.3 0.3500165
1.35 0.38059796
1.4 0.40956308
1.45 0.43723107
1.5 0.46383222
1.55 0.48953882
1.56 0.49458502
1.570796327 0.5
Here is the 5th order quadratic equation driven from that data using mycurvefit.com:
$y = -130.1039 + 489.8105*x - 737.3181*x^2 + 554.5809*x^3 - 208.0569*x^4 + ...$ $31.13006*x^5$
All you do is pop in the ratio $L/c$ for $x$ and then you get $y$, the ratio $h/c$. Since the chord length is known, and you have the ratio $h/c$, you can solve for the height by multiplying the ratio of $h/c$ by the chord length.
...but I kept going...
I wondered if someone reading this might have an application for the inverse segment. Things get less complicated as the ratio of $L/c$ approaches infinity.
Here is the data from $x = pi/2$ to $x = pi$:
x = L/c y = h/c
1.570796327 0.5
1.58 0.50458986
1.59 0.50955019
1.6 0.51448353
1.7 0.56248484
1.8 0.60845528
1.9 0.65281718
2 0.6958726
2.1 0.73784553
2.2 0.778907
2.3 0.81919066
2.4 0.85880306
2.5 0.89783052
2.6 0.93634401
2.7 0.97440257
2.8 1.01205584
2.9 1.04934598
3 1.08630908
3.14159265 1.13814656
This has a polynomial curve fit as follows:
$y = -0.5308871 + 0.8723522*x - 0.1662753*x^2 + 0.01838478*x^3$
Here is the data from $pi$ to $10$:
x = L/c y = h/c
3.14159265 1.13814656
3.25 1.17748049
3.5 1.26717904
3.75 1.35568642
4 1.44321347
4.5 1.61593815
5 1.78627783
5.5 1.9548322
6 2.12201017
7 2.45331872
8 2.78172909
9 3.10812811
10 3.43306996
Which has a polynomial curve fit as follows:
$y = 0.02093882 + 0.364244*x - 0.002337369*x^2$
The relationship between $L/c$ and h/c becomes very linear after this point, as $R^2$ values still are $0.9999$ or better just using a linear equation. Here is the data from $x = 10$ to $100$:
x = L/c y = h/c
10 3.43306996
15 5.04548886
20 6.64810187
25 8.24652077
50 16.21859143
75 24.1813035
100 32.14157177
The linear regression for this curve fit is as follows:
$y = 0.3188628*x + 0.2643024$
Beyond this point, I am not sure if the iterative solver is becoming unreliable due to rounding errors, so use the next information with a grain of salt as it approaches the capabilities of the software. Here is the data from $100$ to $10^{10}$:
x = L/c y = h/c
100 32.14157177
1000 318.627412
100000 31831.30692
10000000000 3183098862
The linear regression is as follows:
$y = 0.3183099*x + 0.3154708$
Despite the growing value of the intercept, it is relatively insignificant in light of the magnitude of the input or output and is probably just a growing rounding error as the software approaches its limitations in the iterative solver. Therefore, beyond $10^{10}$, you can simply use $m = pi^{-1}$ and $b = 0$ in the linear regression formula $y = m*x + b$:
$y = x/pi$
Or in other words, as $x$ approaches infinity, $m$ approaches the inverse of $pi$ and $b$ approaches zero. In practical terms, the smaller the chord is, the closer the sagitta becomes the diameter. I think that passes the smell test.
See this link to download an Excel file solver:
https://drive.google.com/file/d/1er5kIRyL6UahqqAAjlKTEVVhOeFyDZNd/view?usp=sharing
Download it if you want to use it. Don't bother editting or opening in Google Docs. I used features only available in Excel. Once open in Excel, just enter in the Arc and Chord lengths into cells B2 and B3, respectively, and out pops the Sigatta in cell B8.
MAJOR UPDATE
I have since found an additional way to solve this problem thanks to the following resources from 1728.org:
Circle Segment Calculator: http://www.1728.org/circsect.htm
Explanation of Newton's Method: http://www.1728.org/newton.htm
Explained on the second link, I was able to solve for the arc angle using the Newton-Raphson Method of Iterative solving. Usually, the solution converges within 8 decimal places within 4 iterations. However, if the guess is far too great or small, the method fails to find a valid solution (and breaks the solver for the first link). To prevent this, I created a method by which to estimate the initial guess for the half arc angle, mathematically.
$X_{Guess} = pi*(tanh((ln(L/c-1)+0.646543492637008)*0.373118408093598)+1)$
WHERE:
$X_{Guess}$ is an approximation for half the arc angle in radians.
$tanh()$ and $ln()$ are trigonometric and logarithmic functions, respectively.
$L$ is the arc length.
$c$ is the cord length.
The $R^2$ value is about $0.999005$, so this gives a pretty close guess to the arc angle. The largest error I have noticed on my chart is an error value of $2.815% $ as percentage of difference ($X_{Data}$ minus $X_{Guess}$) divided by $pi$. Because it is so close, you can use this as an input to solve for a much closer solution using the Newton-Raphson method as follows:
$f(X_N) = sin(X_N) - c*X_N/L = 0$
$f'(X_N) = cos(X_N) - c/L = 0$
$X_{N+1} = X_N - f(X)/f'(X)$
WHERE:
$f(X_N)$ is an equation of equality given the angle and the chord and arc lengths.
$sin()$ and $cos()$ are trigonometric functions.
$c$ is cord length.
$L$ is the arc length.
$f'(X_N)$ is the derivative of $f(X_N)$.
$X_{N+1}$ is the next iteration of X to solve using the Newton-Raphson Method.
Using the $X_{Guess}$ from the first equation, input it into those three equations to find the next $X_N$ and so on until the $X_N$'s stop changing. For instance, using $L=20$ and $c=18$...
$X_{Guess} = pi*(tanh((ln(20/18-1)+0.646543492637008)*0.373118408093598)+1) = 0.75140903$
$f(X_{Guess}) = sin(0.75140903) - 18*0.75140903/20 = 0.00640093$
$f'(X_{Guess}) = cos(0.75140903) - 18/20 = 0.75140903$
$X_1 = 0.75140903 - 0.00640093/0.75140903 = 0.74289047$
Inputting $X_1$ in place of $X_{Guess}$, $X_2$ in place of $X_1$ and so on...
$X_2 = 0.79071296$
$X_3 = 0.78671238$
$X_4 = 0.78668307$
$X_5 = 0.78668307$
And there we stop, since the solution has converged on 8 decimal places.
Error from guess as a difference ($X_5$ minus $X_{Guess}$) divided by $pi$:
$(0.75140903 - 0.78668307)/pi = -1.123% $
Which, with a 100 FT of material, that 1% could make a difference. Furthermore, this method is arguably more accurate than the first, since it doesn't rely on trend data but instead actual convergence of iterative solving. Plus, that makes it way cooler.
To find the sagitta, you need to first get the radius, which is...
$R = L/(X_F*2)$
WHERE:
$R$ is the radius of the arc.
$L$ is the length of the arc.
$X_F$ is the final iteration of $X_{N+1}$ from the Newton-Raphson Method.
Finally, to find the sagitta from the radius and cord length...
If $L/c < pi/2$, then...
$h = R - sqrt(R^2 - (c/2)^2)$
If $L/c > pi/2$, then...
$h = R + sqrt(R^2 - (c/2)^2)$
WHERE:
$h$ is the sagitta or height of the arc.
$R$ is the arc radius.
$c$ is the cord length of the arc.
Using our example to solve the above equations yields...
$R = 20/(0.78668307*2) = 12.71159932$
Since $0.78668307 < pi/2$ ...
$h = 12.71159932 - sqrt(12.71159932^2 - (18/2)^2) = 3.73469800$
Using the old solver method first mentioned, the resulting sagitta was $3.78510859$ ... Let's solve for the error of the polynomial curve fit vs. Newton-Raphson method as a difference of new minus old over the cord length:
$(3.73469800 - 3.78510859)/18 = -0.28005883% $
So... the old method was not too far off, at least in comparison with my guessing equation. Glad I didn't steer too far off the course from perfect.
To make it all nice and neat, I created a spreadsheet and uploaded it to Google Docs. Here is the link:
https://drive.google.com/file/d/16x5tC17QS8Lj07CMIcRDtiMbe14vnX1d/view?usp=sharing
Download it, and open with Excel. Do not try to open on Google Docs, since I have used features specific to Excel. To use the spreadsheet, type in your arc length and cord length into cells B2 and B3, respectively. The Central Angle, Radius, and Sagitta will be solved in cells B8 (in radians, B9 for degrees), B10, and B11, respectively.
To make this new spreadsheet more hands-free, if the solver method comes up with $X_N=0$ within 16 decimals, the iterations will restart from initial guess, but multiply it by a factor of the number of iterations to the 2.5th power divided by 2.5 ... which, so you know, "2.5" was chosen because for some reason it works very nicely to reduce the number of iterations approaching zero vs. some other values I have tried.
2ND MAJOR UPDATE
Now that I have had some sleep, I realized that the sigmoidal curve I had used for guessing the half arc angle was indeed supposed to be assymetrical. In fact, it is near perfect of an assymetrical sigmoidal curve. It makes sense that my original symmetrical sigmoidal curve was going to have some larger error than necessary.
I was actually shocked to have approximated the answer so closely that it could arguably be used at a larger scale. Certainly, it could be used alone for what I need it for (estimating curve of a fabric). Here is where I have found the information on formulating an equation for an assymetrical sigmoidal curve:
https://www.graphpad.com/guides/prism/7/curve-fitting/index.htm?reg_asymmetric_dose_response_ec.htm
Here is the theoretical curve fit equation for obtaining half of the central arc angle from only the arc length and cord length:
$X = B + (T - B)/(1 + 10^{(E + log_{10}(2^{1/S} - 1)/H - ln(L/c - 1))*H})^S$
WHERE:
$X$ is the half of the central arc angle.
$B$ is the Bottom or lowest possible value of $X$.
$T$ is the Top or highest possible value of $X$.
$E$ is the logEC50 or about where 50% of the concentrations that give half-maximum effects... basically, it moves the curve's inflection point to the left if negative or right if positive.
$log_{10}()$ is a logarithmic function, base $10$.
$S$ is the Symmetry and is a number greater than zero that controls how pinched or loose the upper and lower curves are. Less than one indicates a more gradual curve towards $-infty$ than $+infty$.
$H$ is Hillslope - how steep the slope at the inflection point the curve is.
$ln()$ is a (natural) logarithmic function, base $e$.
$L$ is the arc length.
$c$ is the cord length.
Here are the values for the constants I have determined are most accurate, using Excel's solver tool:
$B = 0$
$T = pi$
$E = -0.559341467113653 cong -0.55934147$
$H = 0.410514378955745 cong 0.41051438$
$S = 0.524651055687417 cong 0.52465106$
This solves to the following equations:
$X = pi/(1 + 10^{(-0.55934147 + log_{10}(2^{1/0.52465106} - 1)/0.41051438 - ln(L/c - 1))*0.41051438})^{0.52465106}$
Math Code (ie., a Google Search):
2*pi/(1+10^((-0.55934147+log(2^(1/0.52465106)-1)/0.41051438-ln(L/c-1))*0.41051438))^0.52465106
Where L and c are replaced by real values.
Excel Code:
2*PI()/(1+10^((-0.55934147+LOG10(2^(1/0.52465106)-1)/0.41051438-LN(B2/B3-1))*0.41051438))^0.52465106
Where L and c are located in cells B2 and B3.
$R = L/(2*pi/(1 + 10^{(-0.55934147 + log_{10}(2^{1/0.52465106} - 1)/0.41051438 - ln(L/c - 1))*0.41051438})^{0.52465106})$
Math Code (ie., a Google Search):
L/(2*pi/(1+10^((-0.55934147+log(2^(1/0.52465106)-1)/0.41051438-ln(L/c-1))*0.41051438))^0.52465106)
Where L and c are replaced by real values.
Excel Code:
B2/(2*PI()/(1+10^((-0.55934147+LOG10(2^(1/0.52465106)-1)/0.41051438-LN(B2/B3-1))*0.41051438))^0.52465106)
Where L and c are located in cells B2 and B3.
If $L/c < pi/2$, Then...
$h = L/(2*pi/(1 + 10^{(-0.55934147 + log_{10}(2^{1/0.52465106} - 1)/0.41051438 - ln(L/c - 1))*0.41051438})^{0.52465106}) - sqrt((L/(2*pi/(1 + 10^{(-0.55934147 + log_{10}(2^{1/0.52465106} - 1)/0.41051438 - ln(L/c - 1))*0.41051438})^{0.52465106}))^2 - (c/2)^2)$
Math Code (ie., a Google Search):
L/(2*pi/(1+10^((-0.55934147+log(2^(1/0.52465106)-1)/0.41051438-ln(L/c-1))*0.41051438))^0.52465106)-sqrt((L/(2*pi/(1+10^((-0.55934147+log(2^(1/0.52465106)-1)/0.41051438-ln(L/c-1))*0.41051438))^0.52465106))^2-(c/2)^2)
Where L and c are replaced by real values.
Excel Code:
B2/(2*PI()/(1+10^((-0.55934147+LOG10(2^(1/0.52465106)-1)/0.41051438-LN(B2/B3-1))*0.41051438))^0.52465106)-SQRT((B2/(2*PI()/(1+10^((-0.55934147+LOG10(2^(1/0.52465106)-1)/0.41051438-LN(B2/B3-1))*0.41051438))^0.52465106))^2-(B3/2)^2)
Where L and c are located in cells B2 and B3.
If $L/c > pi/2$, Then...
$h = L/(2*pi/(1 + 10^{(-0.55934147 + log_{10}(2^{1/0.52465106} - 1)/0.41051438 + ln(L/c - 1))*0.41051438})^{0.52465106}) + sqrt((L/(2*pi/(1 + 10^{(-0.55934147 + log_{10}(2^{1/0.52465106} - 1)/0.41051438 - ln(L/c - 1))*0.41051438})^{0.52465106}))^2 - (c/2)^2)$
Math Code (ie., a Google Search):
L/(2*pi/(1+10^((-0.55934147+log(2^(1/0.52465106)-1)/0.41051438-ln(L/c-1))*0.41051438))^0.52465106)+sqrt((L/(2*pi/(1+10^((-0.55934147+log(2^(1/0.52465106)-1)/0.41051438-ln(L/c-1))*0.41051438))^0.52465106))^2-(c/2)^2)
Where L and c are replaced by real values.
Excel Code:
B2/(2*PI()/(1+10^((-0.55934147+LOG10(2^(1/0.52465106)-1)/0.41051438-LN(B2/B3-1))*0.41051438))^0.52465106)+SQRT((B2/(2*PI()/(1+10^((-0.55934147+LOG10(2^(1/0.52465106)-1)/0.41051438-LN(B2/B3-1))*0.41051438))^0.52465106))^2-(B3/2)^2)
Where L and c are located in cells B2 and B3.
This is quite accurate. For our example up above of $L=20$ and $c=18$, the error as a function of the difference between sagitta guessed vs. solved using the Newton-Raphson method divided by the cord length is only $0.0879% $. The $R^2$ value is a remarkable $0.99999834$ - much better than the first method as a whole. I believe I am the first to come up with this solution for this particular problem (at least publicly on here, since Stigler's law would have you believe different), so just call it the "Devin Approximation" since that name does not seem to be taken and is easier to say than the "Logarithmic Asymmetrical Sigmoidal Approximation".
Here is the Excel file from the first major update now incorporating the new guessing method and having an additional tab called the "Sagitta Solver - Devin Approx." just using the equations above:
https://drive.google.com/file/d/1u5cg56ziQY6_hX9FhWxqFHJjRDAFMGCq/view?usp=sharing
As with the previous files, they must be downloaded and opened in Excel, not to be opened in Google Spreadsheets, since the spreadsheet has features only available in Excel. To use it, enter in the values for arc length and cord length in cells B2 and B3, respectively, and that will solve for the central angle, radius, and sagitta in cells B16, B17, and B18, respectively on the first tab.
answered Dec 12 '18 at 21:47
Devin ThayerDevin Thayer
11
edited Jan 4 at 21:01
answered Dec 12 '18 at 21:47
Devin ThayerDevin Thayer
11
answered Dec 12 '18 at 21:47
Devin ThayerDevin Thayer
11
answered Dec 12 '18 at 21:47
Devin ThayerDevin Thayer
11
11
$begingroup$
Given the effort you put into writing this, it may be worth just a little more effort to put the formulas in MathJax format so people might actually read all of this. You can get started at math.stackexchange.com/help
$endgroup$
– David K
Jan 3 at 19:16
$begingroup$
Oh snap. Thank you. I will get right on it.
$endgroup$
– Devin Thayer
Jan 3 at 19:57
$begingroup$
All set :) fixed.
$endgroup$
– Devin Thayer
Jan 3 at 21:09
add a comment |
$begingroup$
Given the effort you put into writing this, it may be worth just a little more effort to put the formulas in MathJax format so people might actually read all of this. You can get started at math.stackexchange.com/help
$endgroup$
– David K
Jan 3 at 19:16
$begingroup$
Oh snap. Thank you. I will get right on it.
$endgroup$
– Devin Thayer
Jan 3 at 19:57
$begingroup$
All set :) fixed.
$endgroup$
– Devin Thayer
Jan 3 at 21:09
$begingroup$
Given the effort you put into writing this, it may be worth just a little more effort to put the formulas in MathJax format so people might actually read all of this. You can get started at math.stackexchange.com/help
$endgroup$
– David K
Jan 3 at 19:16
$begingroup$
Given the effort you put into writing this, it may be worth just a little more effort to put the formulas in MathJax format so people might actually read all of this. You can get started at math.stackexchange.com/help
$endgroup$
– David K
Jan 3 at 19:16
$begingroup$
Oh snap. Thank you. I will get right on it.
$endgroup$
– Devin Thayer
Jan 3 at 19:57
$begingroup$
Oh snap. Thank you. I will get right on it.
$endgroup$
– Devin Thayer
Jan 3 at 19:57
$begingroup$
All set :) fixed.
$endgroup$
– Devin Thayer
Jan 3 at 21:09
$begingroup$
All set :) fixed.
$endgroup$
– Devin Thayer
Jan 3 at 21:09
add a comment |
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