# Math Help - Parallel Curves Formula

1. ## Parallel Curves Formula

I initially posted this under pre-Claculus, I think that this is a more suitable forum.

I need help using the Parallel Curves formula as I do not understand it.
The Formula is : Parallel curve - Wikipedia, the free encyclopedia

I do not understand how to use it.
I have all points on an initial Ellipse. I want to get a series of points on the parallel ellipse at distance a, so that the resulting Ellipse/Curve is Parallel to my initial one.

I need a numerical example (step by Step) of how to solve this. I am ignorant in differentiation, I am ashamed to say.

The Center of my Ellipse is not at (0,0).

Here are the initial Data I have:
Semi Major Axis = 300
Semi Minor Axis = 100
Center of Ellipse is at (325, 150)
I have the following points that lay on my initial curve (I am only including 10 points for this example)

x=50.0 y=150.0
x=64.60409274419712 y=119.09193189994731
x=105.34840889200964 y=92.11125074884501
x=167.63100531626952 y=70.5919375890824
x=246.84993888980867 y=56.067973463048816
x=338.403266485459 y=50.07333941313367
x=432.12868437393 y=53.79356143875327
x=514.7275697259736 y=66.41038541758431
x=581.4945842792583 y=86.39071388031789
x=627.827784906616 y=112.20056578456459
x=649.1252284808784 y=142.3059600879351

2. Originally Posted by sherif6

I initially posted this under pre-Claculus, I think that this is a more suitable forum.

I need help using the Parallel Curves formula as I do not understand it.
The Formula is : Parallel curve - Wikipedia, the free encyclopedia

I do not understand how to use it.
I have all points on an initial Ellipse. I want to get a series of points on the parallel ellipse at distance a, so that the resulting Ellipse/Curve is Parallel to my initial one.

I need a numerical example (step by Step) of how to solve this. I am ignorant in differentiation, I am ashamed to say.

The Center of my Ellipse is not at (0,0).

Here are the initial Data I have:
Semi Major Axis = 300
Semi Minor Axis = 100
Center of Ellipse is at (325, 150)
I have the following points that lay on my initial curve (I am only including 10 points for this example)

x=50.0 y=150.0
x=64.60409274419712 y=119.09193189994731
x=105.34840889200964 y=92.11125074884501
x=167.63100531626952 y=70.5919375890824
x=246.84993888980867 y=56.067973463048816
x=338.403266485459 y=50.07333941313367
x=432.12868437393 y=53.79356143875327
x=514.7275697259736 y=66.41038541758431
x=581.4945842792583 y=86.39071388031789
x=627.827784906616 y=112.20056578456459
x=649.1252284808784 y=142.3059600879351
Your data doesn't seem to fit the parameters for the ellipse that you have given us. In particular, based on your data points, the center of the ellipse seems to be $C=({\color{red}350},150)$.

Since the relationship between the parameters of the ellipse and your data is not entirely clear, let me avoid pluging in concrete numbers, that might be wrong anyway, and just parametrize your ellipse in the form $x(t)=x_C+a\cos(t), y(t)=y_C+b\sin(t)$, where $a,b$ are the lengths of the semi-axes, and $C=(x_C,y_C)$ the center of the ellipse. This parametrization assumes that the axis of length a is parallel to the x-axis. The parameter t varies from $0$ to $2\pi$.
Given this parametrization, you get that $x'(t)=-a\sin(t), y'(t)=b\cos(t)$, and the parametrization of the parallel curve with distance d to your ellipse comes out to be:

$x(t) = x_C+a\cos(t)+\frac{d\cdot b\cos(t)}{\sqrt{a^2+b^2}}$
$y(t)=y_C+b\sin(t)+\frac{d\cdot a\sin(t)}{\sqrt{a^2+b^2}}$

You are right, the center is at 350. ( the rest of the numbers are correct)
325 is (Length of Major Axis 600 + Offest of 50)/2.... which is not the center.