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Solutions for y for a rotated ellipse

开发者 https://www.devze.com 2023-01-30 20:23 出处:网络
I wish to plot an ellipse by scanline finding the values for y for each value of x. For a plain ellipse the formula is trivial to find: y = Sqrt[b^2 - (b^2 x^2)/a^2]

I wish to plot an ellipse by scanline finding the values for y for each value of x.

For a plain ellipse the formula is trivial to find: y = Sqrt[b^2 - (b^2 x^2)/a^2]

But when th开发者_Go百科e axes of the ellipse are rotated I've never been able to figure out how to compute y (and possibly the extents of x)


In parametric form

x[t]= a Cos[t] Cos[psi] - b Sin[t] Sin[psi]

y[t]= b Cos[psi] Sin[t] + a Cos[t] Sin[psi]

Where psi is the rotation angle, and a and b the semi-axes.

The parameter t goes from 0 to 2 Pi.

Or if you prefer in Cartesian non-parametric form:

(a x^2+b y^2) Cos[psi]^2 + (b x^2 +a y^2) Sin[psi]^2 + (a-b) x y Sin[2 psi]==1

Which yields to the two possible solutions for y[x], equivalent to the two solutions for the square root in the non-rotated case:

y -> (-(Sqrt[2]*Sqrt[a + b - 2*a*b*x^2 + (-a + b)*Cos[2*psi]]) + 
       (-a + b)*x*Sin[2*psi]) / (2*(b*Cos[psi]^2 + a*Sin[psi]^2))

y ->   (Sqrt[2]*Sqrt[a + b - 2*a*b*x^2 + (-a + b)*Cos[2*psi]] + 
       (-a + b)*x*Sin[2*psi])/ (2*(b*Cos[psi]^2 + a*Sin[psi]^2))

Well, you asked for it :)

Those functions give:

Solutions for y for a rotated ellipse

And the limits for x are:

LimitX= +/- Sqrt[a + b + (-a + b)*Cos[2*psi]]/(Sqrt[2]*Sqrt[a]*Sqrt[b])
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