circlesgeometrytrigonometry
It is the end of the day and my brain just can't cope anymore. Can anyone help with, what I hope, is a simple question?
Given a point on a circle ($x$, $y$ coordinates) how can I calculate the coordinates of another point at a given distance around the circumference of the circle? Known variables are the starting point, distance traveled and radius of the circle.
Update 2: Here is the graph I got for $(x_{1},y_{1})=( 78. 965,12. 354)$, for the parametric circle $(x(t),y(t))$ centered at $(100,100)$
$$x=100+90.135\cos \left( 1.3527+\pi -t\frac{\pi }{180}\right) ,$$
$$y=100+90.135\sin \left( 1.3527+\pi -t\frac{\pi }{180}\right) .$$
together with the 4 points $(x(t),y(t))$ for $t=0,90,180,270$
$$(x_{1},y_{1})=(x(0),y(0)),(x(90),y(90)),(x(180),y(180)),(x(270),y(270)).$$
You might use the following equations in a for loop with $k=0$ to $k=359$, step $1$:
$$x=100+90.135\cos \left( 1.3527+\pi -k\frac{\pi }{180}\right) ,$$
$$y=100+90.135\sin \left( 1.3527+\pi -k\frac{\pi }{180}\right) .$$
to draw the "orbit" with a 1 degree interval.
Update: corrected coordinates of $(x_{1},y_{1})=(140.5,152)$.
You need to consider the new angle and not only the $1{{}^\circ}$ change. The argument of $\cos$ and $\sin$ is this new angle and not $1{{}^\circ}$.
Let $(x_{c},y_{c})=(160,240)$ be the center of the set of circles and $(x_{1},y_{1})=(140.5,152)$. The radius $r$ is
$$\begin{eqnarray*} r &=&\sqrt{\left( x_{c}-x_{1}\right) ^{2}+\left( y_{c}-y_{1}\right) ^{2}} \\ &=&\sqrt{\left( 160-140.5\right) ^{2}+\left( 240-152\right) ^{2}} \\ &=&90.135 \end{eqnarray*}$$
Call $(x,y)$ the new coordinates of $(x_{1},y_{1})$ rotated by an angle of $-1{{}^\circ}=-\dfrac{\pi }{180}$ around $(x_{c},y_{c})$ with a radius $r$. The new angle is $\theta'=\theta -\frac{\pi }{180}$, $\theta $ being the initial angle. Then
$$\begin{eqnarray*} x &=&x_{c}+r\cos \left( \theta -\frac{\pi }{180}\right), \\ y &=&y_{c}+r\sin \left( \theta -\frac{\pi }{180}\right), \end{eqnarray*}$$ where $\theta $ is the angle $\theta =\arctan \dfrac{y_{1}-y_{c}}{x_{1}-x_{c}}:$
$$\begin{eqnarray*} \theta &=&\arctan \frac{152-240}{140.5-160}=1.3527+\pi \text{ rad.}\\ &=&\frac{1.3527\times 180{{}^\circ}}{\pi }+180{{}^\circ}=257. 5{{}^\circ}\end{eqnarray*}$$ Thus $$\begin{eqnarray*} x &=&160+90.135\cos \left( 1.3527+\pi -\frac{\pi }{180}\right)= 138. 96 \\y &=&240+90.135\sin \left( 1.3527+\pi -\frac{\pi }{180}\right) = 152. 35 \end{eqnarray*}$$
You can derive a simple formula using the law of cosines. In fact, while all the planar geometry is helpful for visualization, there's really no need for most of it. You have 3 points: your arc start and stop points, which I'll call $A$ and $B$, and your circle center, $C$. The angle for the arc you're wanting to measure, I'll call it $\theta$, is the angle of the triangle $ABC$ at point $C$. Because $C$ is the center of the circle that $A$ and $B$ are on, the triangle sides $AC$ and $BC$ are equal to your circle's radius, $r$. We'll call the length of $AB$, the remaining side, $d$ (see picture).
(By the way, if $A=(x_1,y_1)$ and $B=(x_2,y_2)$, then $d=\sqrt {(x_1-x_2)^2 + (y_1-y_2)^2}$.)
According to the law of cosines, $\cos (\theta )={{r^2+r^2-d^2}\over {2rr}}=1- {{d^2}\over {2r^2}}$.
So all you need is the distance between the end points of your arc and the radius of the circle to compute the angle,
$\theta = \arccos (1- {{d^2}\over {2r^2}})$
Lastly, the length is calculated -
$Length = r\theta$
Where $\theta$ is expressed in radians.
Best Answer
Let the center of the circle be $(a,b)$. Drag the center to the origin. This procedure drags our given point $(x,y)$ to $(x-a,y-b)$.
For simplicity of notation, let $u=x-a$, $v=y-b$. Now we determine the angle that the positive $x$-axis has to be rotated through (counterclockwise) to hit the line from the origin to $(u,v)$. Call this angle $\theta$.
Then $\theta$ is the angle, say in the interval $(-\pi,\pi]$, whose cosine is $u/r$, and whose sine is $v/r$, where $r$ is the radius of the circle. (This was already known; it also happens to be $\sqrt{u^2+v^2}$.) So from now on we can take $\theta$ as known. But we have to be careful to take the signs of $u$ and $v$ into account when calculating $\theta$.
Let $D$ be the distance travelled. Assume that we are travelling counterclockwise. Then the angle of travel is (in radians) equal to $D/r$. Let $\phi$ be this angle. If we are travelling clockwise, just replace $\phi$ by $-\phi$. So from now on we can take $\phi$ as known.
After the travel, our angle is $\theta+\phi$. This means that we are at the point with coordinates
$$(r\cos(\theta+\phi), \: \: r\sin(\theta+\phi)).$$
Now transform back, by adding $(a,b)$ to the point. We obtain $$(a+r\cos(\theta+\phi),\:\: b+r\sin(\theta+\phi)).$$
All the components of this formula are known, so we can compute the answer.
Comment: Note that $$\cos(\theta+\phi)=\cos\theta\cos\phi-\sin\theta\sin\phi.$$ But we know that $r\cos\theta=u$ and $r\sin\theta=v$. Thus $$a+r\cos(\theta+\phi)=a+u\cos\phi-v\sin\phi.$$
Similarly, $$\sin(\theta+\phi)=\cos\theta\sin\phi+\sin\theta\cos\phi,$$ and therefore $$b+r\sin(\theta+\phi)=u\sin\phi+v\cos\phi.$$ Thus an alternate (and for many purposes simpler) version of the answer is $$(a+(x-a)\cos\phi-(y-b)\sin\phi,\:\: b+(x-a)\sin\phi+(y-b)\cos\phi),$$ where $\phi=D/r$.
We could also reach this by quoting the rotation formula. Recall that when we are rotating a point $(u,v)$ about the origin through an angle $\phi$, we multiply the vector by a certain matrix. You can think of this post as being, in particular, a derivation of the rotation formula.