Mixing
Circle circumference point calculation
$begingroup$
In the image below, I have a part of a circle. Given,
$$text{chord }d=1050 mm\
text{height }f=50 mm\
text{radius }R=2781 mm\
text{centre }O(700 mm,2781 mm)$$
and $3$ points $A(0,200), B(1050,200), C(525,150)$.
I would like to know if there is an equation to calculate the $y$ coordinate for any given $x$ coordinate, for example the point $D(200,Y)$.
trigonometry circle
edited Jan 14 at 13:32
Shubham Johri
5,192717
asked Jan 14 at 13:11
AleksanderAleksander
11
$endgroup$
add a comment |
$begingroup$
In the image below, I have a part of a circle. Given,
$$text{chord }d=1050 mm\
text{height }f=50 mm\
text{radius }R=2781 mm\
text{centre }O(700 mm,2781 mm)$$
and $3$ points $A(0,200), B(1050,200), C(525,150)$.
I would like to know if there is an equation to calculate the $y$ coordinate for any given $x$ coordinate, for example the point $D(200,Y)$.
trigonometry circle
edited Jan 14 at 13:32
Shubham Johri
5,192717
asked Jan 14 at 13:11
AleksanderAleksander
11
$endgroup$
$begingroup$
I would start by figuring out the equation of the circle. I mean, in the form $$ (x-x_0)^2 + (y-y_0)^2 = R^2 $$
$endgroup$
– Matti P.
Jan 14 at 13:20
$begingroup$
The question has already been asked here.
$endgroup$
– whiskeyo
Jan 14 at 13:22
add a comment |
$begingroup$
In the image below, I have a part of a circle. Given,
$$text{chord }d=1050 mm\
text{height }f=50 mm\
text{radius }R=2781 mm\
text{centre }O(700 mm,2781 mm)$$
and $3$ points $A(0,200), B(1050,200), C(525,150)$.
I would like to know if there is an equation to calculate the $y$ coordinate for any given $x$ coordinate, for example the point $D(200,Y)$.
trigonometry circle
edited Jan 14 at 13:32
Shubham Johri
5,192717
asked Jan 14 at 13:11
AleksanderAleksander
11
$endgroup$
In the image below, I have a part of a circle. Given,
$$text{chord }d=1050 mm\
text{height }f=50 mm\
text{radius }R=2781 mm\
text{centre }O(700 mm,2781 mm)$$
and $3$ points $A(0,200), B(1050,200), C(525,150)$.
I would like to know if there is an equation to calculate the $y$ coordinate for any given $x$ coordinate, for example the point $D(200,Y)$.
trigonometry circle
trigonometry circle
edited Jan 14 at 13:32
Shubham Johri
5,192717
asked Jan 14 at 13:11
AleksanderAleksander
11
edited Jan 14 at 13:32
Shubham Johri
5,192717
asked Jan 14 at 13:11
AleksanderAleksander
11
edited Jan 14 at 13:32
Shubham Johri
5,192717
edited Jan 14 at 13:32
Shubham Johri
5,192717
edited Jan 14 at 13:32
Shubham Johri
5,192717
5,192717
asked Jan 14 at 13:11
AleksanderAleksander
11
asked Jan 14 at 13:11
AleksanderAleksander
11
asked Jan 14 at 13:11
AleksanderAleksander
11
11
$begingroup$
I would start by figuring out the equation of the circle. I mean, in the form $$ (x-x_0)^2 + (y-y_0)^2 = R^2 $$
$endgroup$
– Matti P.
Jan 14 at 13:20
$begingroup$
The question has already been asked here.
$endgroup$
– whiskeyo
Jan 14 at 13:22
add a comment |
$begingroup$
I would start by figuring out the equation of the circle. I mean, in the form $$ (x-x_0)^2 + (y-y_0)^2 = R^2 $$
$endgroup$
– Matti P.
Jan 14 at 13:20
$begingroup$
The question has already been asked here.
$endgroup$
– whiskeyo
Jan 14 at 13:22
$begingroup$
I would start by figuring out the equation of the circle. I mean, in the form $$ (x-x_0)^2 + (y-y_0)^2 = R^2 $$
$endgroup$
– Matti P.
Jan 14 at 13:20
$begingroup$
I would start by figuring out the equation of the circle. I mean, in the form $$ (x-x_0)^2 + (y-y_0)^2 = R^2 $$
$endgroup$
– Matti P.
Jan 14 at 13:20
$begingroup$
The question has already been asked here.
$endgroup$
– whiskeyo
Jan 14 at 13:22
$begingroup$
The question has already been asked here.
$endgroup$
– whiskeyo
Jan 14 at 13:22
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Try to solve something more general: Suppose that you have 3 arbitrary points $A(x_A,y_A)$, $B(x_B,y_B)$ and $C(x_C,y_C)$. Let's find center $O(x_O,y_O)$ and radius $R$ of the circle passing through all three points.
Denote midpoints of segments $AB$,$BC$ with $P$,$Q$:
$$P(x_P=frac{x_A+x_B}{2}, y_P=frac{y_A+x_B}{2})$$
$$Q(x_Q=frac{x_B+x_C}{2}, y_Q=frac{y_B+y_C}{2})$$
Obviously $OPbot AB$ and $OQbot BC$ which means that:
$$(x_O-x_P)(x_B-x_A)+(y_O-y_P)(y_B-y_A)=0$$
$$(x_O-x_Q)(x_C-x_B)+(y_O-y_Q)(y_C-y_B)=0$$
or:
$$(x_O-frac{x_A+x_B}{2})(x_B-x_A)+(y_O-frac{y_A+y_B}{2})(y_B-y_A)=0$$
$$(x_O-frac{x_B+x_C}{2})(x_C-x_B)+(y_O-frac{y_B+y_C}{2})(y_C-y_B)=0$$
or:
$$(x_B-x_A)x_O+(y_B-y_A)y_O=frac{x_B^2-x_A^2+y_B^2-y_A^2}{2}tag{1}$$
$$(x_C-x_B)x_O+(y_C-y_B)y_O=frac{x_C^2-x_B^2+y_C^2-y_B^2}{2}tag{2}$$
The point is: equations (1) and (2) are linear wtih two unknowns $(x_O,y_O)$ and, assuming that points $A,B,C$ are not collinear, you can always find a unique solution. You can also come up with a general formula for $(x_O,y_O)$ but I leave it up to you as an exercise.
Once you have $x_O,y_O$ you can find the radius of the circle:
$$R=sqrt{(x_A-x_O)^2+(y_A-y_O)^2}$$
For any point on the circle $K(x_K,y_K)$:
$$(x_K-x_O)^2+(y_K-y_O)^2=R^2$$
And if you know $x_K$, you can calculate the value of $y_K$:
$$y_K=y_Opmsqrt{R^2-(x_K-x_O)^2}$$
answered Jan 14 at 14:43
OldboyOldboy
8,1651936
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Try to solve something more general: Suppose that you have 3 arbitrary points $A(x_A,y_A)$, $B(x_B,y_B)$ and $C(x_C,y_C)$. Let's find center $O(x_O,y_O)$ and radius $R$ of the circle passing through all three points.
Denote midpoints of segments $AB$,$BC$ with $P$,$Q$:
$$P(x_P=frac{x_A+x_B}{2}, y_P=frac{y_A+x_B}{2})$$
$$Q(x_Q=frac{x_B+x_C}{2}, y_Q=frac{y_B+y_C}{2})$$
Obviously $OPbot AB$ and $OQbot BC$ which means that:
$$(x_O-x_P)(x_B-x_A)+(y_O-y_P)(y_B-y_A)=0$$
$$(x_O-x_Q)(x_C-x_B)+(y_O-y_Q)(y_C-y_B)=0$$
or:
$$(x_O-frac{x_A+x_B}{2})(x_B-x_A)+(y_O-frac{y_A+y_B}{2})(y_B-y_A)=0$$
$$(x_O-frac{x_B+x_C}{2})(x_C-x_B)+(y_O-frac{y_B+y_C}{2})(y_C-y_B)=0$$
or:
$$(x_B-x_A)x_O+(y_B-y_A)y_O=frac{x_B^2-x_A^2+y_B^2-y_A^2}{2}tag{1}$$
$$(x_C-x_B)x_O+(y_C-y_B)y_O=frac{x_C^2-x_B^2+y_C^2-y_B^2}{2}tag{2}$$
The point is: equations (1) and (2) are linear wtih two unknowns $(x_O,y_O)$ and, assuming that points $A,B,C$ are not collinear, you can always find a unique solution. You can also come up with a general formula for $(x_O,y_O)$ but I leave it up to you as an exercise.
Once you have $x_O,y_O$ you can find the radius of the circle:
$$R=sqrt{(x_A-x_O)^2+(y_A-y_O)^2}$$
For any point on the circle $K(x_K,y_K)$:
$$(x_K-x_O)^2+(y_K-y_O)^2=R^2$$
And if you know $x_K$, you can calculate the value of $y_K$:
$$y_K=y_Opmsqrt{R^2-(x_K-x_O)^2}$$
answered Jan 14 at 14:43
OldboyOldboy
8,1651936
$endgroup$
add a comment |
$begingroup$
Try to solve something more general: Suppose that you have 3 arbitrary points $A(x_A,y_A)$, $B(x_B,y_B)$ and $C(x_C,y_C)$. Let's find center $O(x_O,y_O)$ and radius $R$ of the circle passing through all three points.
Denote midpoints of segments $AB$,$BC$ with $P$,$Q$:
$$P(x_P=frac{x_A+x_B}{2}, y_P=frac{y_A+x_B}{2})$$
$$Q(x_Q=frac{x_B+x_C}{2}, y_Q=frac{y_B+y_C}{2})$$
Obviously $OPbot AB$ and $OQbot BC$ which means that:
$$(x_O-x_P)(x_B-x_A)+(y_O-y_P)(y_B-y_A)=0$$
$$(x_O-x_Q)(x_C-x_B)+(y_O-y_Q)(y_C-y_B)=0$$
or:
$$(x_O-frac{x_A+x_B}{2})(x_B-x_A)+(y_O-frac{y_A+y_B}{2})(y_B-y_A)=0$$
$$(x_O-frac{x_B+x_C}{2})(x_C-x_B)+(y_O-frac{y_B+y_C}{2})(y_C-y_B)=0$$
or:
$$(x_B-x_A)x_O+(y_B-y_A)y_O=frac{x_B^2-x_A^2+y_B^2-y_A^2}{2}tag{1}$$
$$(x_C-x_B)x_O+(y_C-y_B)y_O=frac{x_C^2-x_B^2+y_C^2-y_B^2}{2}tag{2}$$
The point is: equations (1) and (2) are linear wtih two unknowns $(x_O,y_O)$ and, assuming that points $A,B,C$ are not collinear, you can always find a unique solution. You can also come up with a general formula for $(x_O,y_O)$ but I leave it up to you as an exercise.
Once you have $x_O,y_O$ you can find the radius of the circle:
$$R=sqrt{(x_A-x_O)^2+(y_A-y_O)^2}$$
For any point on the circle $K(x_K,y_K)$:
$$(x_K-x_O)^2+(y_K-y_O)^2=R^2$$
And if you know $x_K$, you can calculate the value of $y_K$:
$$y_K=y_Opmsqrt{R^2-(x_K-x_O)^2}$$
answered Jan 14 at 14:43
OldboyOldboy
8,1651936
$endgroup$
add a comment |
$begingroup$
Try to solve something more general: Suppose that you have 3 arbitrary points $A(x_A,y_A)$, $B(x_B,y_B)$ and $C(x_C,y_C)$. Let's find center $O(x_O,y_O)$ and radius $R$ of the circle passing through all three points.
Denote midpoints of segments $AB$,$BC$ with $P$,$Q$:
$$P(x_P=frac{x_A+x_B}{2}, y_P=frac{y_A+x_B}{2})$$
$$Q(x_Q=frac{x_B+x_C}{2}, y_Q=frac{y_B+y_C}{2})$$
Obviously $OPbot AB$ and $OQbot BC$ which means that:
$$(x_O-x_P)(x_B-x_A)+(y_O-y_P)(y_B-y_A)=0$$
$$(x_O-x_Q)(x_C-x_B)+(y_O-y_Q)(y_C-y_B)=0$$
or:
$$(x_O-frac{x_A+x_B}{2})(x_B-x_A)+(y_O-frac{y_A+y_B}{2})(y_B-y_A)=0$$
$$(x_O-frac{x_B+x_C}{2})(x_C-x_B)+(y_O-frac{y_B+y_C}{2})(y_C-y_B)=0$$
or:
$$(x_B-x_A)x_O+(y_B-y_A)y_O=frac{x_B^2-x_A^2+y_B^2-y_A^2}{2}tag{1}$$
$$(x_C-x_B)x_O+(y_C-y_B)y_O=frac{x_C^2-x_B^2+y_C^2-y_B^2}{2}tag{2}$$
The point is: equations (1) and (2) are linear wtih two unknowns $(x_O,y_O)$ and, assuming that points $A,B,C$ are not collinear, you can always find a unique solution. You can also come up with a general formula for $(x_O,y_O)$ but I leave it up to you as an exercise.
Once you have $x_O,y_O$ you can find the radius of the circle:
$$R=sqrt{(x_A-x_O)^2+(y_A-y_O)^2}$$
For any point on the circle $K(x_K,y_K)$:
$$(x_K-x_O)^2+(y_K-y_O)^2=R^2$$
And if you know $x_K$, you can calculate the value of $y_K$:
$$y_K=y_Opmsqrt{R^2-(x_K-x_O)^2}$$
answered Jan 14 at 14:43
OldboyOldboy
8,1651936
$endgroup$
Try to solve something more general: Suppose that you have 3 arbitrary points $A(x_A,y_A)$, $B(x_B,y_B)$ and $C(x_C,y_C)$. Let's find center $O(x_O,y_O)$ and radius $R$ of the circle passing through all three points.
Denote midpoints of segments $AB$,$BC$ with $P$,$Q$:
$$P(x_P=frac{x_A+x_B}{2}, y_P=frac{y_A+x_B}{2})$$
$$Q(x_Q=frac{x_B+x_C}{2}, y_Q=frac{y_B+y_C}{2})$$
Obviously $OPbot AB$ and $OQbot BC$ which means that:
$$(x_O-x_P)(x_B-x_A)+(y_O-y_P)(y_B-y_A)=0$$
$$(x_O-x_Q)(x_C-x_B)+(y_O-y_Q)(y_C-y_B)=0$$
or:
$$(x_O-frac{x_A+x_B}{2})(x_B-x_A)+(y_O-frac{y_A+y_B}{2})(y_B-y_A)=0$$
$$(x_O-frac{x_B+x_C}{2})(x_C-x_B)+(y_O-frac{y_B+y_C}{2})(y_C-y_B)=0$$
or:
$$(x_B-x_A)x_O+(y_B-y_A)y_O=frac{x_B^2-x_A^2+y_B^2-y_A^2}{2}tag{1}$$
$$(x_C-x_B)x_O+(y_C-y_B)y_O=frac{x_C^2-x_B^2+y_C^2-y_B^2}{2}tag{2}$$
The point is: equations (1) and (2) are linear wtih two unknowns $(x_O,y_O)$ and, assuming that points $A,B,C$ are not collinear, you can always find a unique solution. You can also come up with a general formula for $(x_O,y_O)$ but I leave it up to you as an exercise.
Once you have $x_O,y_O$ you can find the radius of the circle:
$$R=sqrt{(x_A-x_O)^2+(y_A-y_O)^2}$$
For any point on the circle $K(x_K,y_K)$:
$$(x_K-x_O)^2+(y_K-y_O)^2=R^2$$
And if you know $x_K$, you can calculate the value of $y_K$:
$$y_K=y_Opmsqrt{R^2-(x_K-x_O)^2}$$
answered Jan 14 at 14:43
OldboyOldboy
8,1651936
answered Jan 14 at 14:43
OldboyOldboy
8,1651936
answered Jan 14 at 14:43
OldboyOldboy
8,1651936
answered Jan 14 at 14:43
OldboyOldboy
8,1651936
8,1651936
add a comment |
add a comment |
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$begingroup$
I would start by figuring out the equation of the circle. I mean, in the form $$ (x-x_0)^2 + (y-y_0)^2 = R^2 $$
$endgroup$
– Matti P.
Jan 14 at 13:20
$begingroup$
The question has already been asked here.
$endgroup$
– whiskeyo
Jan 14 at 13:22