0

I have an Ellipse with known coordinates , I would like to know how I can create an inner ellipse coordinates that are exactly 5 inches perpendicular from the outer ellipse points. Please see the drawing at the link - I have the outer points x,y posted below. Axis Major a 15.75, axis minor b 12.

    X     ,  Y
-6.0000,-0.6250
-6.0000,-1.2500
-6.0000,-1.8750
-6.0000,-2.5000
-5.9800,-3.1073
-5.8900,-3.7091
-5.7400,-4.2990
-5.5400,-4.8712
-5.2800,-5.4197
-4.9600,-5.9388
-4.5900,-6.4233
-4.1800,-6.8682
-3.7200,-7.2689
-3.2300,-7.6213
-2.7000,-7.9218
-2.1400,-8.1673
-1.5600,-8.3553
-0.9700,-8.4838
-0.3700,-8.5516
0.3700,-8.5516
0.9700,-8.4838
1.5600,-8.3553
2.1400,-8.1673
2.7000,-7.9218
3.2300,-7.6213
3.7200,-7.2689
4.1800,-6.8682
4.5900,-6.4233
4.9600,-5.9388
5.2800,-5.4197
5.5400,-4.8712
5.7400,-4.2990
5.8900,-3.7091
5.9800,-3.1073
6.0000,-2.5000
6.0000,-1.8750
6.0000,-1.2500
6.0000,-0.6250

DRAWING LOCATED HERE : Inner ellipse generated from points on outer ellipse.

ADDED PLOT USING DOUG M formula: Excel Scatter Plot ]2 This is somewhat off as the inner Ellipse should match the outer only smaller by 5 ..

EXCEL 
cellX =A5-5 * ( (A5/15.75^2) / (SQRT(( (A5^2) / (15.75^4) ) + ( (B5^2) / (12^4) ))))

cellY =B5-5 * ( (B5/12^2) / (SQRT(( (A5^2) / (15.75^4) ) + ( (B5^2) / (12^4) ))))

where:
A = X values.
B = Y values. 
axis major a = 15.75  I changed the semi to 8.5 as it matches the ellipse.
axis minor b = 12 - Semi = 6

EDIT - 4/18/2016

Attached is a plot in Excel : While these points are 5 inches away from the outer ellipse - they are not perpendicular to the points on the ellipse, I need the perpendicular point -5 from the outer ellipse. Dougs equation does a good job of calculating 5 points away but maybe it will not plot to be an ellipse - which will work for me as long as the points are perpendicular to the outer ellipse. I am unable to add the additional plot - as I do not have my account on this site combined to the other SO site (same account).

StixO
  • 105

1 Answers1

0

$\frac {x^2}{a^2} + \frac {y^2}{b^2} = 1$

Suppose we look at

$z = \frac {x^2}{a^2} + \frac {y^2}{b^2} \\ \nabla z = \frac {2x}{a^2}, \frac {2y}{b^2} $

This vector is perpendicular to the elipse.

if $(x,y)$ is on your elipse, then $(x,y) - 5\dfrac{(\frac {x}{a^2},\frac {y}{b^2})}{\sqrt{\frac {x^2}{a^4}+\frac {y^2}{b^4}}}$ is on the curve you are looking for.

Doug M
  • 57,877
  • does this also work if this is considered a super ellipse ? The math seems straight forward . I will take this and try to apply it. I am not sure how to apply the equation - given the comma in the numerator. Is there a link that would explain this formula - how to apply it ? – StixO Apr 15 '16 at 02:42
  • In the equation at the commas am I really just substituting which to solve first? – StixO Apr 15 '16 at 13:31
  • the comma separates the x coordinate from the y coordinate. – Doug M Apr 15 '16 at 15:13
  • @ Doug M - I understand the separation but I guess I do not understand the solving of the problem for example if I have 2 values x,y 1,2 placing in 1/a^2 and using that as my numerator to solve for my x point (of course leaving the denominator with both x and y values. Then for y substitute the 2 in for y and my numerator would be made up of only the y/b^2 component ? – StixO Apr 15 '16 at 18:27
  • I plotted the points in Excel and attached the snapshot , I listed the formula in the problem edit. The points are close but do not follow the outer as I should have the outer ellipse duplicated only smaller (5 points). Is there a way to do this? – StixO Apr 15 '16 at 20:51
  • a and b are the lengths of the semi-major axis and semi-minor axis. – Doug M Apr 15 '16 at 21:22
  • The plot is 5 away but the points are not perpendicular with the ellipse points - perhaps the actual result will not be an ellipse but may look more like a bulb thermometer in the end. Perhaps I need to post the question differently. BTW: thanks for the formula - I have learned something. – StixO Apr 18 '16 at 15:31