Perimeter of a part of an ellipse

I know the ellipse(center of ellipse,length of semi-axes,orientation).I also know the co-ordinates of two points which comprise the end points of the curve which is a part of the given ellipse.How can I find out the perimeter of this part of an ellipse??

[Torsten]

You know how to compute the arc length of a curve in parametric form ?

http://www.mathwords.com/a/arc_length_of_a_curve.htm

Best wishes

Torsten.

Answer 1

[John BG]

Sibananda

If you already have a and b, then you don't really need the ellipse centre point, or the ellipse tilt angle.

The points of the ellipse have a unique angle, let's say the pair of points you want to measure the section of perimeter in between are

p1: angle1

p2: angle2

then the length you are after

L=perim_ellipse(angle1,a,b)-perim_ellipse(angle2,a,b)

where perim_ellipse is a function that calculates the perimeter length from 0 to angle.

If you really know it's an ellipse, or what's the same: you can afford neglecting the approximation error then you don't really need the parametric representation of the ellipse and you may <http://mathworld.wolfram.com/Ellipse.html>

Perimeter=a*E(t,sqrt(1-b^2/a^2))

should coincide with

b*E(t,sqrt(1-a^2/b^2))

where E(t,k) is an elliptic integral

for instance

a=1
b=2

k1=sqrt(1-b^2/a^2)
fun1=@(angle) sqrt(1-k1^2*(sin(angle)).^2)
E1=integral(fun1,0,2*pi)

E1 =
   9.688448220547675

k2=sqrt(1-a^2/b^2)
fun2=@(angle) sqrt(1-k2^2*(sin(angle)).^2)
E2=integral(fun2,0,2*pi)
E2 =
   4.844224110273838

and 

   E1/E2
   =
       2

2.- go straight to the S.Ramanujan 2nd approximation:

a=1;b=2;  % for instance
h=(a-b)^2/(a+b)^2
h =
   0.111111111111111

   Perimeter=pi*(a+b)*(1+3*h/(10+(4-3*h)))
  Perimeter =
     9.654650593958877

E1-Perimeter
 =
   0.033797626588798

more approximations here:

http://www.numericana.com/answer/ellipse.htm#elliptic http://www.johndcook.com/blog/2013/05/05/ramanujan-circumference-ellipse/ http://paulbourke.net/geometry/ellipsecirc/

2.- if you actually have the points, then is when the parametric formulation of the ellipse can be integrated,

Centring and removing tilt the parametric equations of an ellipse with axes a and be are:

t=[0:.01:2*pi]
x=a*cos(t)
y=b*sin(t)

L=[x;y];
perimeter=0;
for k=2:1:length(t) 
    perimeter=perimeter+dist1(L(:,k-1),L(:,k)); 
end

perimeter =
   9.682037252420420

So,

function L1=perim_ellipse(t1,a,b)
% t1 angle in radian of ellipse point
% a,b: ellipse absolute minor and major axes, without tilt, in any order
t=[0:.01:t1];
  x=a*cos(t);
  y=b*sin(t);
  ellipse_points=[x;y];
  L1=0;
  for k=2:1:length(t) 
      L1=L1+dist1(ellipse_points(:,k-1),ellipse_points(:,k)); 
  end
end

checking

format long;
L0=perim_ellipse(2*pi,1,2);
L0 =
   9.682037252420420

So, to calculate the arc length you asked, of let's say angle1=45º and angle2=135º:

angle1=pi/4;
angle2=3*pi/4;
a=1;b=2;
L12=abs(perim_ellipse(angle1,a,b)-perim_ellipse(angle2,a,b))
L12 =
   1.930084466978536

If you find this answer of any help solving your question,

please click on the thumbs-up vote link, or mark this answer as accepted

thanks in advance

John

[e-mail]

[SIBANANDA PANIGRAHY]

Thank you very much for your answer.

[John BG]

would you like to mark my answer as accepted? regards

John

[e-mail]

Answer 2

[Roger Stafford]

If (x0,y0) is the center of the ellipse, if a and b are the two semi-axis lengths, and if p is the counterclockwise angle of the a-semi-axis orientation with respect the the x-axis, then the entire ellipse can be represented parametrically by the equations

   x = x0 + a*cos(p)*cos(t) - b*sin(p)*sin(t)
   y = y0 + a*sin(p)*cos(t) + b*cos(p)*sin(t)

where t is a parameter varying from -pi to +pi. If (x1,y1) and (x2,y2) are the two endpoints of the elliptical arc in question in such a way that the desired arc goes counterclockwise from (x1,y1) to (x2,y2), then the corresponding endpoint parameters t1 and t2 can be solved for as follows:

   A = [a*cos(p),-b*sin(p);a*sin(p),b*cos(p)];
   T1 = A\[x1-x0;y1-y0]; % Solve for cos(t1) and sin(t1)
   t1 = atan2(T1(2),T1(1)); % Determine t1
   T2 = A\[x2-x0;y2-y0]; % Solve for cos(t2) and sin(t2)
   t2 = atan2(T2(2),T2(1)); % Determine t2
   if t2 < t1, t2 = t2 + 2*pi; end % Ensure that t2 >= t1
   t = linspace(t1,t2); % Make t start at t1 and end at t2
   x = x0 + a*cos(p)*cos(t) - b*sin(p)*sin(t);
   y = y0 + a*sin(p)*cos(t) + b*cos(p)*sin(t);

The x and y vectors will trace the desired elliptical arc. To verify this with a plot, you can do this:

   tt = linspace(-pi,pi);
   xx = x0 + a*cos(p)*cos(tt) - b*sin(p)*sin(tt);
   yy = y0 + a*sin(p)*cos(tt) + b*cos(p)*sin(tt);
   plot(xx,yy,'c:',x0,y0,'w*',x,y,'y-',x1,y1,'ro',x2,y2,'go')
   axis equal

The desired arc is in yellow, the remainder of the ellipse is in dotted cyan, the point (x1,y1) is red, (x2,y2) is green, and the center of the ellipse is a white asterisk.

[Roger Stafford]

Where you said ‘perimeter’, Sibananda, it now seems apparent that you meant perimeter arclength for the elliptical curve segment between the two points. If that is the case, this arclength can be found, as Torsten indicated, taking the integral with respect to the parameter t of the expression "sqrt((dx/dt)^2+(dy/dt)^2)" from the t1 to the t2 determined above. In this case this expression will be equal to

   sqrt(a^2*sin(t)^2+b^2*cos(t)^2)

and thus the desired arclength would be:

   s = integral(@(t) sqrt(a^2*sin(t).^2+b^2*cos(t).^2),t1,t2); % Corrected

(This integral can be also be evaluated in terms of elliptic integrals of the second kind, incomplete.)