Mixing
Diagonal calculation in a 3D square
$begingroup$
I have the information as shown in the following image:
i.e.: all sides (AB, BC CD and DA) and one diagonal (BD) and the height difference between the point A and the points B/C/D (855.35mm).
How do I calculate the length of diagonal AC (red line)?
euclidean-geometry triangle 3d
edited Jan 11 at 21:19
user376343
3,7383827
asked Jan 11 at 14:36
A.BaworA.Bawor
81
$endgroup$
add a comment |
$begingroup$
I have the information as shown in the following image:
i.e.: all sides (AB, BC CD and DA) and one diagonal (BD) and the height difference between the point A and the points B/C/D (855.35mm).
How do I calculate the length of diagonal AC (red line)?
euclidean-geometry triangle 3d
edited Jan 11 at 21:19
user376343
3,7383827
asked Jan 11 at 14:36
A.BaworA.Bawor
81
$endgroup$
$begingroup$
That’s not really a “square,” now, is it? The figure isn’t even planar. You do have a triangular “base,” though, so try starting by working out the relative coordinates of those three points.
$endgroup$
– amd
Jan 11 at 23:16
$begingroup$
There are two possible locations for $A$ that have different distances from $C$, so you need some other criteria for choosing the correct one.
$endgroup$
– amd
Jan 12 at 0:28
add a comment |
$begingroup$
I have the information as shown in the following image:
i.e.: all sides (AB, BC CD and DA) and one diagonal (BD) and the height difference between the point A and the points B/C/D (855.35mm).
How do I calculate the length of diagonal AC (red line)?
euclidean-geometry triangle 3d
edited Jan 11 at 21:19
user376343
3,7383827
asked Jan 11 at 14:36
A.BaworA.Bawor
81
$endgroup$
I have the information as shown in the following image:
i.e.: all sides (AB, BC CD and DA) and one diagonal (BD) and the height difference between the point A and the points B/C/D (855.35mm).
How do I calculate the length of diagonal AC (red line)?
euclidean-geometry triangle 3d
euclidean-geometry triangle 3d
edited Jan 11 at 21:19
user376343
3,7383827
asked Jan 11 at 14:36
A.BaworA.Bawor
81
edited Jan 11 at 21:19
user376343
3,7383827
asked Jan 11 at 14:36
A.BaworA.Bawor
81
edited Jan 11 at 21:19
user376343
3,7383827
edited Jan 11 at 21:19
user376343
3,7383827
edited Jan 11 at 21:19
user376343
3,7383827
3,7383827
asked Jan 11 at 14:36
A.BaworA.Bawor
81
asked Jan 11 at 14:36
A.BaworA.Bawor
81
asked Jan 11 at 14:36
A.BaworA.Bawor
81
81
$begingroup$
That’s not really a “square,” now, is it? The figure isn’t even planar. You do have a triangular “base,” though, so try starting by working out the relative coordinates of those three points.
$endgroup$
– amd
Jan 11 at 23:16
$begingroup$
There are two possible locations for $A$ that have different distances from $C$, so you need some other criteria for choosing the correct one.
$endgroup$
– amd
Jan 12 at 0:28
add a comment |
$begingroup$
That’s not really a “square,” now, is it? The figure isn’t even planar. You do have a triangular “base,” though, so try starting by working out the relative coordinates of those three points.
$endgroup$
– amd
Jan 11 at 23:16
$begingroup$
There are two possible locations for $A$ that have different distances from $C$, so you need some other criteria for choosing the correct one.
$endgroup$
– amd
Jan 12 at 0:28
$begingroup$
That’s not really a “square,” now, is it? The figure isn’t even planar. You do have a triangular “base,” though, so try starting by working out the relative coordinates of those three points.
$endgroup$
– amd
Jan 11 at 23:16
$begingroup$
That’s not really a “square,” now, is it? The figure isn’t even planar. You do have a triangular “base,” though, so try starting by working out the relative coordinates of those three points.
$endgroup$
– amd
Jan 11 at 23:16
$begingroup$
There are two possible locations for $A$ that have different distances from $C$, so you need some other criteria for choosing the correct one.
$endgroup$
– amd
Jan 12 at 0:28
$begingroup$
There are two possible locations for $A$ that have different distances from $C$, so you need some other criteria for choosing the correct one.
$endgroup$
– amd
Jan 12 at 0:28
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It is not a square but a tetrahedron with base $BCD$ and height $h = 855.35$.
Label the edge lengths as
$$begin{cases}
b &= |CD| = 2578,\
c &= |BD| = 2828,\
d &= |BC| = 2128,\
b_1 &= |AB| = 2060,\
c_1 &= |AC|\
d_1 &= |AD| = 2045\
end{cases}$$
$c_1$ will be the length we seek.
You can compute the volume of this tetrahedron in two different manners.
$V = frac13 mathcal{A} h$ where $mathcal{A}$ is area of $triangle BCD$.
You can use Heron's formula to get $mathcal{A}$.
$$16mathcal{A}^2 = (b+c+d)(-b+c+d)(b-c+d)(b+c-d)$$You can also compute the volume using Cayley Menger determinant.
$$288V^2 = left|begin{matrix}
0 & 1 & 1 & 1 & 1\
1 & 0 & b_1^2 & c_1^2 & d_1^2\
1 & b_1^2 & 0 & d^2 & c^2\
1 & c_1^2 & d^2 & 0 & b^2\
1 & d_1^2 & c^2 & b^2 & 0
end{matrix}right|$$
Expanding the CM determinant and combine it with first result, one obtain following equation which is quadratic in $c_1^2$:
$$begin{align}
16mathcal{A}^2h^2 = 144V^2
=& phantom{+}; b^2 b_1^2(-b^2-b_1^2 + c^2 + color{red}{c_1^2} + d^2 + d_1^2)\
& + c^2 color{red}{c_1^2}(;b^2+b_1^2 - c^2 - color{red}{c_1^2} + d^2 + d_1^2)\
& + d^2 d_1^2(;b^2+b_1^2 + c^2 + color{red}{c_1^2} - d^2 - d_1^2)\
& - (b^2c^2d^2 + b^2 color{red}{c_1^2} d_1^2 + b_1^2 c^2 d_1^2 + b_1^2color{red}{c_1^2} d^2)
end{align}$$
With help of a CAS, we can substitute back the numerical values of $b,c,d,b_1,d_1$ into above equation and simplify. The end result is
$$begin{align}c_1
&= sqrt{frac{116153047144695pmsqrt{8168336037304042557755678133}}{19993960}}\
&approx 1135.385089196282 ;text{ or }; 3213.987289241557
end{align}
$$
There are two possible solutions for $c_1$. The '+' solution corresponds to the
case where the dihedral angle between the planes holding $triangle ABD$ and $triangle BCD$ is obtuse ( $> 90^circ$). For the '-' solution, the dihedral angle is acute ($< 90^circ$).
Judging from your picture, the dihedral angle at edge $BD$ is obtuse. The length you seek is the one $approx 3213.987289241557$.
answered Jan 12 at 6:10
achille huiachille hui
95.9k5132258
$endgroup$
$begingroup$
I am trying to create an excel spreadsheet to calculate c1, but I am not sure how to get the quadratic equation from the 5x5 matrix. Is this even possible in a excel? Perhaps I should just use the quadratic equation for c1, but again I don't know what it is. Could you expand on that please?
$endgroup$
– A.Bawor
Jan 12 at 19:21
$begingroup$
Thank you for elaborating on this, but I'll be honest. I unfortunately still can't figure out how to simplify this equation, because as far as I know excel cannot multiply numbers with letters i.e.: 15*c will not become 15c. Therefore if I substitute the numbers back into this equation, then it will not work as I have c1 as unknown, so it gives me an error. Apologies for being too much of a newbie...
$endgroup$
– A.Bawor
Jan 13 at 15:21
$begingroup$
@A.Bawor You can always use the Goal seek feature in Excel. (I'm using Excel 2007, it is accessible through the menuData->Data tools->What-if analysis) 1. Setup 7 cells for the values of $b,c,d,b_1,c_1,d_1,h$ (put a guess value to $c_1$). 2. Place the entries of CM matrix in $5 times 5$ cells. 3. Select the $5 times 5$ cells, right click to name the range asCM4. create a cell and use the function=MDETERM(CM)to compute the determinant of the named range CM. 5. create another cell which compute the difference of this determinant with $32mathcal{A}^2 h^2$.
$endgroup$
– achille hui
Jan 13 at 16:50
$begingroup$
6. Use the Goal seek feature to find the $c_1$ value which make the cell in step 5 goes to zero.
$endgroup$
– achille hui
Jan 13 at 16:50
$begingroup$
Thank you so much for all of your help! Works like a charm ;)
$endgroup$
– A.Bawor
Jan 13 at 19:03
add a comment |
$begingroup$
Judging from the fact that A = (?, ?, 855.35), I think the figure is a tetrahedron instead of a square. It has BCD as the base and AA’ (= 855.35) as its vertical height. Therefore, all arrow-marked angles shown are $90^0$. Also, AC is a slant edge instead of a diagonal.
Apply Pythagoras to [$triangle ABA’$ with AB and AA’ known] to get BA’.
Similarly, DA’ can be found from $triangle ADA'$.
Apply Cosine law to [$triangle BDC$ with all 3 sides known] to get $angle BDC$.
Similarly, apply Cosine law to [$triangle BDA’$ with all 3 sides known] to get $angle BDA’$.
$angle A’DC = angle BDC - angle BDA’$
Then, apply Cosine law [$triangle A’DC$ with DA’, DC and $angle A’DC$ known to get A’C.
Finally, apply Pythagoras to [$triangle AA’C$] to get AC.
answered Jan 12 at 3:23
MickMick
11.8k21641
$endgroup$
add a comment |
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2 Answers
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2 Answers
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oldest
votes
$begingroup$
It is not a square but a tetrahedron with base $BCD$ and height $h = 855.35$.
Label the edge lengths as
$$begin{cases}
b &= |CD| = 2578,\
c &= |BD| = 2828,\
d &= |BC| = 2128,\
b_1 &= |AB| = 2060,\
c_1 &= |AC|\
d_1 &= |AD| = 2045\
end{cases}$$
$c_1$ will be the length we seek.
You can compute the volume of this tetrahedron in two different manners.
$V = frac13 mathcal{A} h$ where $mathcal{A}$ is area of $triangle BCD$.
You can use Heron's formula to get $mathcal{A}$.
$$16mathcal{A}^2 = (b+c+d)(-b+c+d)(b-c+d)(b+c-d)$$You can also compute the volume using Cayley Menger determinant.
$$288V^2 = left|begin{matrix}
0 & 1 & 1 & 1 & 1\
1 & 0 & b_1^2 & c_1^2 & d_1^2\
1 & b_1^2 & 0 & d^2 & c^2\
1 & c_1^2 & d^2 & 0 & b^2\
1 & d_1^2 & c^2 & b^2 & 0
end{matrix}right|$$
Expanding the CM determinant and combine it with first result, one obtain following equation which is quadratic in $c_1^2$:
$$begin{align}
16mathcal{A}^2h^2 = 144V^2
=& phantom{+}; b^2 b_1^2(-b^2-b_1^2 + c^2 + color{red}{c_1^2} + d^2 + d_1^2)\
& + c^2 color{red}{c_1^2}(;b^2+b_1^2 - c^2 - color{red}{c_1^2} + d^2 + d_1^2)\
& + d^2 d_1^2(;b^2+b_1^2 + c^2 + color{red}{c_1^2} - d^2 - d_1^2)\
& - (b^2c^2d^2 + b^2 color{red}{c_1^2} d_1^2 + b_1^2 c^2 d_1^2 + b_1^2color{red}{c_1^2} d^2)
end{align}$$
With help of a CAS, we can substitute back the numerical values of $b,c,d,b_1,d_1$ into above equation and simplify. The end result is
$$begin{align}c_1
&= sqrt{frac{116153047144695pmsqrt{8168336037304042557755678133}}{19993960}}\
&approx 1135.385089196282 ;text{ or }; 3213.987289241557
end{align}
$$
There are two possible solutions for $c_1$. The '+' solution corresponds to the
case where the dihedral angle between the planes holding $triangle ABD$ and $triangle BCD$ is obtuse ( $> 90^circ$). For the '-' solution, the dihedral angle is acute ($< 90^circ$).
Judging from your picture, the dihedral angle at edge $BD$ is obtuse. The length you seek is the one $approx 3213.987289241557$.
answered Jan 12 at 6:10
achille huiachille hui
95.9k5132258
$endgroup$
$begingroup$
I am trying to create an excel spreadsheet to calculate c1, but I am not sure how to get the quadratic equation from the 5x5 matrix. Is this even possible in a excel? Perhaps I should just use the quadratic equation for c1, but again I don't know what it is. Could you expand on that please?
$endgroup$
– A.Bawor
Jan 12 at 19:21
$begingroup$
Thank you for elaborating on this, but I'll be honest. I unfortunately still can't figure out how to simplify this equation, because as far as I know excel cannot multiply numbers with letters i.e.: 15*c will not become 15c. Therefore if I substitute the numbers back into this equation, then it will not work as I have c1 as unknown, so it gives me an error. Apologies for being too much of a newbie...
$endgroup$
– A.Bawor
Jan 13 at 15:21
$begingroup$
@A.Bawor You can always use the Goal seek feature in Excel. (I'm using Excel 2007, it is accessible through the menuData->Data tools->What-if analysis) 1. Setup 7 cells for the values of $b,c,d,b_1,c_1,d_1,h$ (put a guess value to $c_1$). 2. Place the entries of CM matrix in $5 times 5$ cells. 3. Select the $5 times 5$ cells, right click to name the range asCM4. create a cell and use the function=MDETERM(CM)to compute the determinant of the named range CM. 5. create another cell which compute the difference of this determinant with $32mathcal{A}^2 h^2$.
$endgroup$
– achille hui
Jan 13 at 16:50
$begingroup$
6. Use the Goal seek feature to find the $c_1$ value which make the cell in step 5 goes to zero.
$endgroup$
– achille hui
Jan 13 at 16:50
$begingroup$
Thank you so much for all of your help! Works like a charm ;)
$endgroup$
– A.Bawor
Jan 13 at 19:03
add a comment |
$begingroup$
It is not a square but a tetrahedron with base $BCD$ and height $h = 855.35$.
Label the edge lengths as
$$begin{cases}
b &= |CD| = 2578,\
c &= |BD| = 2828,\
d &= |BC| = 2128,\
b_1 &= |AB| = 2060,\
c_1 &= |AC|\
d_1 &= |AD| = 2045\
end{cases}$$
$c_1$ will be the length we seek.
You can compute the volume of this tetrahedron in two different manners.
$V = frac13 mathcal{A} h$ where $mathcal{A}$ is area of $triangle BCD$.
You can use Heron's formula to get $mathcal{A}$.
$$16mathcal{A}^2 = (b+c+d)(-b+c+d)(b-c+d)(b+c-d)$$You can also compute the volume using Cayley Menger determinant.
$$288V^2 = left|begin{matrix}
0 & 1 & 1 & 1 & 1\
1 & 0 & b_1^2 & c_1^2 & d_1^2\
1 & b_1^2 & 0 & d^2 & c^2\
1 & c_1^2 & d^2 & 0 & b^2\
1 & d_1^2 & c^2 & b^2 & 0
end{matrix}right|$$
Expanding the CM determinant and combine it with first result, one obtain following equation which is quadratic in $c_1^2$:
$$begin{align}
16mathcal{A}^2h^2 = 144V^2
=& phantom{+}; b^2 b_1^2(-b^2-b_1^2 + c^2 + color{red}{c_1^2} + d^2 + d_1^2)\
& + c^2 color{red}{c_1^2}(;b^2+b_1^2 - c^2 - color{red}{c_1^2} + d^2 + d_1^2)\
& + d^2 d_1^2(;b^2+b_1^2 + c^2 + color{red}{c_1^2} - d^2 - d_1^2)\
& - (b^2c^2d^2 + b^2 color{red}{c_1^2} d_1^2 + b_1^2 c^2 d_1^2 + b_1^2color{red}{c_1^2} d^2)
end{align}$$
With help of a CAS, we can substitute back the numerical values of $b,c,d,b_1,d_1$ into above equation and simplify. The end result is
$$begin{align}c_1
&= sqrt{frac{116153047144695pmsqrt{8168336037304042557755678133}}{19993960}}\
&approx 1135.385089196282 ;text{ or }; 3213.987289241557
end{align}
$$
There are two possible solutions for $c_1$. The '+' solution corresponds to the
case where the dihedral angle between the planes holding $triangle ABD$ and $triangle BCD$ is obtuse ( $> 90^circ$). For the '-' solution, the dihedral angle is acute ($< 90^circ$).
Judging from your picture, the dihedral angle at edge $BD$ is obtuse. The length you seek is the one $approx 3213.987289241557$.
answered Jan 12 at 6:10
achille huiachille hui
95.9k5132258
$endgroup$
$begingroup$
I am trying to create an excel spreadsheet to calculate c1, but I am not sure how to get the quadratic equation from the 5x5 matrix. Is this even possible in a excel? Perhaps I should just use the quadratic equation for c1, but again I don't know what it is. Could you expand on that please?
$endgroup$
– A.Bawor
Jan 12 at 19:21
$begingroup$
Thank you for elaborating on this, but I'll be honest. I unfortunately still can't figure out how to simplify this equation, because as far as I know excel cannot multiply numbers with letters i.e.: 15*c will not become 15c. Therefore if I substitute the numbers back into this equation, then it will not work as I have c1 as unknown, so it gives me an error. Apologies for being too much of a newbie...
$endgroup$
– A.Bawor
Jan 13 at 15:21
$begingroup$
@A.Bawor You can always use the Goal seek feature in Excel. (I'm using Excel 2007, it is accessible through the menuData->Data tools->What-if analysis) 1. Setup 7 cells for the values of $b,c,d,b_1,c_1,d_1,h$ (put a guess value to $c_1$). 2. Place the entries of CM matrix in $5 times 5$ cells. 3. Select the $5 times 5$ cells, right click to name the range asCM4. create a cell and use the function=MDETERM(CM)to compute the determinant of the named range CM. 5. create another cell which compute the difference of this determinant with $32mathcal{A}^2 h^2$.
$endgroup$
– achille hui
Jan 13 at 16:50
$begingroup$
6. Use the Goal seek feature to find the $c_1$ value which make the cell in step 5 goes to zero.
$endgroup$
– achille hui
Jan 13 at 16:50
$begingroup$
Thank you so much for all of your help! Works like a charm ;)
$endgroup$
– A.Bawor
Jan 13 at 19:03
add a comment |
$begingroup$
It is not a square but a tetrahedron with base $BCD$ and height $h = 855.35$.
Label the edge lengths as
$$begin{cases}
b &= |CD| = 2578,\
c &= |BD| = 2828,\
d &= |BC| = 2128,\
b_1 &= |AB| = 2060,\
c_1 &= |AC|\
d_1 &= |AD| = 2045\
end{cases}$$
$c_1$ will be the length we seek.
You can compute the volume of this tetrahedron in two different manners.
$V = frac13 mathcal{A} h$ where $mathcal{A}$ is area of $triangle BCD$.
You can use Heron's formula to get $mathcal{A}$.
$$16mathcal{A}^2 = (b+c+d)(-b+c+d)(b-c+d)(b+c-d)$$You can also compute the volume using Cayley Menger determinant.
$$288V^2 = left|begin{matrix}
0 & 1 & 1 & 1 & 1\
1 & 0 & b_1^2 & c_1^2 & d_1^2\
1 & b_1^2 & 0 & d^2 & c^2\
1 & c_1^2 & d^2 & 0 & b^2\
1 & d_1^2 & c^2 & b^2 & 0
end{matrix}right|$$
Expanding the CM determinant and combine it with first result, one obtain following equation which is quadratic in $c_1^2$:
$$begin{align}
16mathcal{A}^2h^2 = 144V^2
=& phantom{+}; b^2 b_1^2(-b^2-b_1^2 + c^2 + color{red}{c_1^2} + d^2 + d_1^2)\
& + c^2 color{red}{c_1^2}(;b^2+b_1^2 - c^2 - color{red}{c_1^2} + d^2 + d_1^2)\
& + d^2 d_1^2(;b^2+b_1^2 + c^2 + color{red}{c_1^2} - d^2 - d_1^2)\
& - (b^2c^2d^2 + b^2 color{red}{c_1^2} d_1^2 + b_1^2 c^2 d_1^2 + b_1^2color{red}{c_1^2} d^2)
end{align}$$
With help of a CAS, we can substitute back the numerical values of $b,c,d,b_1,d_1$ into above equation and simplify. The end result is
$$begin{align}c_1
&= sqrt{frac{116153047144695pmsqrt{8168336037304042557755678133}}{19993960}}\
&approx 1135.385089196282 ;text{ or }; 3213.987289241557
end{align}
$$
There are two possible solutions for $c_1$. The '+' solution corresponds to the
case where the dihedral angle between the planes holding $triangle ABD$ and $triangle BCD$ is obtuse ( $> 90^circ$). For the '-' solution, the dihedral angle is acute ($< 90^circ$).
Judging from your picture, the dihedral angle at edge $BD$ is obtuse. The length you seek is the one $approx 3213.987289241557$.
answered Jan 12 at 6:10
achille huiachille hui
95.9k5132258
$endgroup$
It is not a square but a tetrahedron with base $BCD$ and height $h = 855.35$.
Label the edge lengths as
$$begin{cases}
b &= |CD| = 2578,\
c &= |BD| = 2828,\
d &= |BC| = 2128,\
b_1 &= |AB| = 2060,\
c_1 &= |AC|\
d_1 &= |AD| = 2045\
end{cases}$$
$c_1$ will be the length we seek.
You can compute the volume of this tetrahedron in two different manners.
$V = frac13 mathcal{A} h$ where $mathcal{A}$ is area of $triangle BCD$.
You can use Heron's formula to get $mathcal{A}$.
$$16mathcal{A}^2 = (b+c+d)(-b+c+d)(b-c+d)(b+c-d)$$You can also compute the volume using Cayley Menger determinant.
$$288V^2 = left|begin{matrix}
0 & 1 & 1 & 1 & 1\
1 & 0 & b_1^2 & c_1^2 & d_1^2\
1 & b_1^2 & 0 & d^2 & c^2\
1 & c_1^2 & d^2 & 0 & b^2\
1 & d_1^2 & c^2 & b^2 & 0
end{matrix}right|$$
Expanding the CM determinant and combine it with first result, one obtain following equation which is quadratic in $c_1^2$:
$$begin{align}
16mathcal{A}^2h^2 = 144V^2
=& phantom{+}; b^2 b_1^2(-b^2-b_1^2 + c^2 + color{red}{c_1^2} + d^2 + d_1^2)\
& + c^2 color{red}{c_1^2}(;b^2+b_1^2 - c^2 - color{red}{c_1^2} + d^2 + d_1^2)\
& + d^2 d_1^2(;b^2+b_1^2 + c^2 + color{red}{c_1^2} - d^2 - d_1^2)\
& - (b^2c^2d^2 + b^2 color{red}{c_1^2} d_1^2 + b_1^2 c^2 d_1^2 + b_1^2color{red}{c_1^2} d^2)
end{align}$$
With help of a CAS, we can substitute back the numerical values of $b,c,d,b_1,d_1$ into above equation and simplify. The end result is
$$begin{align}c_1
&= sqrt{frac{116153047144695pmsqrt{8168336037304042557755678133}}{19993960}}\
&approx 1135.385089196282 ;text{ or }; 3213.987289241557
end{align}
$$
There are two possible solutions for $c_1$. The '+' solution corresponds to the
case where the dihedral angle between the planes holding $triangle ABD$ and $triangle BCD$ is obtuse ( $> 90^circ$). For the '-' solution, the dihedral angle is acute ($< 90^circ$).
Judging from your picture, the dihedral angle at edge $BD$ is obtuse. The length you seek is the one $approx 3213.987289241557$.
answered Jan 12 at 6:10
achille huiachille hui
95.9k5132258
edited Jan 12 at 20:43
answered Jan 12 at 6:10
achille huiachille hui
95.9k5132258
answered Jan 12 at 6:10
achille huiachille hui
95.9k5132258
answered Jan 12 at 6:10
achille huiachille hui
95.9k5132258
95.9k5132258
$begingroup$
I am trying to create an excel spreadsheet to calculate c1, but I am not sure how to get the quadratic equation from the 5x5 matrix. Is this even possible in a excel? Perhaps I should just use the quadratic equation for c1, but again I don't know what it is. Could you expand on that please?
$endgroup$
– A.Bawor
Jan 12 at 19:21
$begingroup$
Thank you for elaborating on this, but I'll be honest. I unfortunately still can't figure out how to simplify this equation, because as far as I know excel cannot multiply numbers with letters i.e.: 15*c will not become 15c. Therefore if I substitute the numbers back into this equation, then it will not work as I have c1 as unknown, so it gives me an error. Apologies for being too much of a newbie...
$endgroup$
– A.Bawor
Jan 13 at 15:21
$begingroup$
@A.Bawor You can always use the Goal seek feature in Excel. (I'm using Excel 2007, it is accessible through the menuData->Data tools->What-if analysis) 1. Setup 7 cells for the values of $b,c,d,b_1,c_1,d_1,h$ (put a guess value to $c_1$). 2. Place the entries of CM matrix in $5 times 5$ cells. 3. Select the $5 times 5$ cells, right click to name the range asCM4. create a cell and use the function=MDETERM(CM)to compute the determinant of the named range CM. 5. create another cell which compute the difference of this determinant with $32mathcal{A}^2 h^2$.
$endgroup$
– achille hui
Jan 13 at 16:50
$begingroup$
6. Use the Goal seek feature to find the $c_1$ value which make the cell in step 5 goes to zero.
$endgroup$
– achille hui
Jan 13 at 16:50
$begingroup$
Thank you so much for all of your help! Works like a charm ;)
$endgroup$
– A.Bawor
Jan 13 at 19:03
add a comment |
$begingroup$
I am trying to create an excel spreadsheet to calculate c1, but I am not sure how to get the quadratic equation from the 5x5 matrix. Is this even possible in a excel? Perhaps I should just use the quadratic equation for c1, but again I don't know what it is. Could you expand on that please?
$endgroup$
– A.Bawor
Jan 12 at 19:21
$begingroup$
Thank you for elaborating on this, but I'll be honest. I unfortunately still can't figure out how to simplify this equation, because as far as I know excel cannot multiply numbers with letters i.e.: 15*c will not become 15c. Therefore if I substitute the numbers back into this equation, then it will not work as I have c1 as unknown, so it gives me an error. Apologies for being too much of a newbie...
$endgroup$
– A.Bawor
Jan 13 at 15:21
$begingroup$
@A.Bawor You can always use the Goal seek feature in Excel. (I'm using Excel 2007, it is accessible through the menuData->Data tools->What-if analysis) 1. Setup 7 cells for the values of $b,c,d,b_1,c_1,d_1,h$ (put a guess value to $c_1$). 2. Place the entries of CM matrix in $5 times 5$ cells. 3. Select the $5 times 5$ cells, right click to name the range asCM4. create a cell and use the function=MDETERM(CM)to compute the determinant of the named range CM. 5. create another cell which compute the difference of this determinant with $32mathcal{A}^2 h^2$.
$endgroup$
– achille hui
Jan 13 at 16:50
$begingroup$
6. Use the Goal seek feature to find the $c_1$ value which make the cell in step 5 goes to zero.
$endgroup$
– achille hui
Jan 13 at 16:50
$begingroup$
Thank you so much for all of your help! Works like a charm ;)
$endgroup$
– A.Bawor
Jan 13 at 19:03
$begingroup$
I am trying to create an excel spreadsheet to calculate c1, but I am not sure how to get the quadratic equation from the 5x5 matrix. Is this even possible in a excel? Perhaps I should just use the quadratic equation for c1, but again I don't know what it is. Could you expand on that please?
$endgroup$
– A.Bawor
Jan 12 at 19:21
$begingroup$
I am trying to create an excel spreadsheet to calculate c1, but I am not sure how to get the quadratic equation from the 5x5 matrix. Is this even possible in a excel? Perhaps I should just use the quadratic equation for c1, but again I don't know what it is. Could you expand on that please?
$endgroup$
– A.Bawor
Jan 12 at 19:21
$begingroup$
Thank you for elaborating on this, but I'll be honest. I unfortunately still can't figure out how to simplify this equation, because as far as I know excel cannot multiply numbers with letters i.e.: 15*c will not become 15c. Therefore if I substitute the numbers back into this equation, then it will not work as I have c1 as unknown, so it gives me an error. Apologies for being too much of a newbie...
$endgroup$
– A.Bawor
Jan 13 at 15:21
$begingroup$
Thank you for elaborating on this, but I'll be honest. I unfortunately still can't figure out how to simplify this equation, because as far as I know excel cannot multiply numbers with letters i.e.: 15*c will not become 15c. Therefore if I substitute the numbers back into this equation, then it will not work as I have c1 as unknown, so it gives me an error. Apologies for being too much of a newbie...
$endgroup$
– A.Bawor
Jan 13 at 15:21
$begingroup$
@A.Bawor You can always use the Goal seek feature in Excel. (I'm using Excel 2007, it is accessible through the menu
Data -> Data tools -> What-if analysis) 1. Setup 7 cells for the values of $b,c,d,b_1,c_1,d_1,h$ (put a guess value to $c_1$). 2. Place the entries of CM matrix in $5 times 5$ cells. 3. Select the $5 times 5$ cells, right click to name the range as CM 4. create a cell and use the function =MDETERM(CM) to compute the determinant of the named range CM. 5. create another cell which compute the difference of this determinant with $32mathcal{A}^2 h^2$.$endgroup$
– achille hui
Jan 13 at 16:50
$begingroup$
@A.Bawor You can always use the Goal seek feature in Excel. (I'm using Excel 2007, it is accessible through the menu
Data -> Data tools -> What-if analysis) 1. Setup 7 cells for the values of $b,c,d,b_1,c_1,d_1,h$ (put a guess value to $c_1$). 2. Place the entries of CM matrix in $5 times 5$ cells. 3. Select the $5 times 5$ cells, right click to name the range as CM 4. create a cell and use the function =MDETERM(CM) to compute the determinant of the named range CM. 5. create another cell which compute the difference of this determinant with $32mathcal{A}^2 h^2$.$endgroup$
– achille hui
Jan 13 at 16:50
$begingroup$
6. Use the Goal seek feature to find the $c_1$ value which make the cell in step 5 goes to zero.
$endgroup$
– achille hui
Jan 13 at 16:50
$begingroup$
6. Use the Goal seek feature to find the $c_1$ value which make the cell in step 5 goes to zero.
$endgroup$
– achille hui
Jan 13 at 16:50
$begingroup$
Thank you so much for all of your help! Works like a charm ;)
$endgroup$
– A.Bawor
Jan 13 at 19:03
$begingroup$
Thank you so much for all of your help! Works like a charm ;)
$endgroup$
– A.Bawor
Jan 13 at 19:03
add a comment |
$begingroup$
Judging from the fact that A = (?, ?, 855.35), I think the figure is a tetrahedron instead of a square. It has BCD as the base and AA’ (= 855.35) as its vertical height. Therefore, all arrow-marked angles shown are $90^0$. Also, AC is a slant edge instead of a diagonal.
Apply Pythagoras to [$triangle ABA’$ with AB and AA’ known] to get BA’.
Similarly, DA’ can be found from $triangle ADA'$.
Apply Cosine law to [$triangle BDC$ with all 3 sides known] to get $angle BDC$.
Similarly, apply Cosine law to [$triangle BDA’$ with all 3 sides known] to get $angle BDA’$.
$angle A’DC = angle BDC - angle BDA’$
Then, apply Cosine law [$triangle A’DC$ with DA’, DC and $angle A’DC$ known to get A’C.
Finally, apply Pythagoras to [$triangle AA’C$] to get AC.
answered Jan 12 at 3:23
MickMick
11.8k21641
$endgroup$
add a comment |
$begingroup$
Judging from the fact that A = (?, ?, 855.35), I think the figure is a tetrahedron instead of a square. It has BCD as the base and AA’ (= 855.35) as its vertical height. Therefore, all arrow-marked angles shown are $90^0$. Also, AC is a slant edge instead of a diagonal.
Apply Pythagoras to [$triangle ABA’$ with AB and AA’ known] to get BA’.
Similarly, DA’ can be found from $triangle ADA'$.
Apply Cosine law to [$triangle BDC$ with all 3 sides known] to get $angle BDC$.
Similarly, apply Cosine law to [$triangle BDA’$ with all 3 sides known] to get $angle BDA’$.
$angle A’DC = angle BDC - angle BDA’$
Then, apply Cosine law [$triangle A’DC$ with DA’, DC and $angle A’DC$ known to get A’C.
Finally, apply Pythagoras to [$triangle AA’C$] to get AC.
answered Jan 12 at 3:23
MickMick
11.8k21641
$endgroup$
add a comment |
$begingroup$
Judging from the fact that A = (?, ?, 855.35), I think the figure is a tetrahedron instead of a square. It has BCD as the base and AA’ (= 855.35) as its vertical height. Therefore, all arrow-marked angles shown are $90^0$. Also, AC is a slant edge instead of a diagonal.
Apply Pythagoras to [$triangle ABA’$ with AB and AA’ known] to get BA’.
Similarly, DA’ can be found from $triangle ADA'$.
Apply Cosine law to [$triangle BDC$ with all 3 sides known] to get $angle BDC$.
Similarly, apply Cosine law to [$triangle BDA’$ with all 3 sides known] to get $angle BDA’$.
$angle A’DC = angle BDC - angle BDA’$
Then, apply Cosine law [$triangle A’DC$ with DA’, DC and $angle A’DC$ known to get A’C.
Finally, apply Pythagoras to [$triangle AA’C$] to get AC.
answered Jan 12 at 3:23
MickMick
11.8k21641
$endgroup$
Judging from the fact that A = (?, ?, 855.35), I think the figure is a tetrahedron instead of a square. It has BCD as the base and AA’ (= 855.35) as its vertical height. Therefore, all arrow-marked angles shown are $90^0$. Also, AC is a slant edge instead of a diagonal.
Apply Pythagoras to [$triangle ABA’$ with AB and AA’ known] to get BA’.
Similarly, DA’ can be found from $triangle ADA'$.
Apply Cosine law to [$triangle BDC$ with all 3 sides known] to get $angle BDC$.
Similarly, apply Cosine law to [$triangle BDA’$ with all 3 sides known] to get $angle BDA’$.
$angle A’DC = angle BDC - angle BDA’$
Then, apply Cosine law [$triangle A’DC$ with DA’, DC and $angle A’DC$ known to get A’C.
Finally, apply Pythagoras to [$triangle AA’C$] to get AC.
answered Jan 12 at 3:23
MickMick
11.8k21641
edited Jan 12 at 3:28
answered Jan 12 at 3:23
MickMick
11.8k21641
answered Jan 12 at 3:23
MickMick
11.8k21641
answered Jan 12 at 3:23
MickMick
11.8k21641
11.8k21641
add a comment |
add a comment |
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$begingroup$
That’s not really a “square,” now, is it? The figure isn’t even planar. You do have a triangular “base,” though, so try starting by working out the relative coordinates of those three points.
$endgroup$
– amd
Jan 11 at 23:16
$begingroup$
There are two possible locations for $A$ that have different distances from $C$, so you need some other criteria for choosing the correct one.
$endgroup$
– amd
Jan 12 at 0:28