For an elliptical spot, I want to calculate the average diameter (or radius). As you can see in the picture, the diameter depends on the coordinate system chosen. If you rotate the coordinate system, then the diameter obtained are different. In the picture these diameters are [a,b] and [x,y] --> depending on the coordinate system chosen.
Now if I take average for these two set of diameters, which way will be a better choice? The arithmetic mean changes, while rotating the coordinate system. So, a+b is not equal to x+y. Does the RMS diameter remains constant with respect to the coordinate system? Is, a^2 + b^2 = x^2 + y^2 ?
Dear Prasenjit Roy
The relation between x2 + y2 and a2 + b2
Following the notations in the attached image:
The Sum x²+y²
In fact as the new axis y = (tanθ)x and y = (tan(θ+(π/2)))x
where θ is the angle of rotation.
The equation of the Ellipse is
(x/a)² + (y/b)² = 1... (E)
(y=(tanθ)x)∩(E) =(m,n)
we get ( x/a)²+((tanθ)x)/b))²=1→x²(1/(a²) +((tanθ)²)/b²) = 1 →
m² = (1/(a²tan²θ+b²))a²b²→
m = (ab)/(√(a²tan²θ+b²))), n=((tanθ)ab)/(√(a²tan²θ+b²))
x²= m2 + n2 = (ab/(√(a²tan²θ+b²))²+((tanθ)ab)/(√(a²tan²θ+b²))² = a²b²((tan²θ+1)/(a²tan²θ+b²))= a²b²(((secθ)²)/(a²tan²θ+b²))
x = ((secθ)ab)/(√(a²tan²θ+b²)))=((ab)/((cosθ)√(a²tan²θ+b²)))
similarly, determine (y = (tan(θ+(π/2)))x)∩(E) = (s, t)
we obtain y = sqrt(s2 + t2 )= (((sec(θ+(π/2)))ab)/(√(a²tan²(θ+(π/2))+b²)))= (ab)/((sinθ)√(a²cot²θ+b²)).
Finally, we calculate
x² + y² = ((ab/(√(a²sin²θ+b²cos²θ))²+( ab/(√(a²cos²θ+b²sin²θ))² = a²b²(a²+b²))/((b²cos²θ+a²sin²θ)(a²cos²θ+b²sin²θ).
Concerning your question:
The answer is given by the following remark:
Remark: x² + y² = a² + b² if and only if θ = kπ or θ = (π/2)+ kπ where
k is any integer and θ is the angle of rotation.
Apply the above calculations to find any term as AM and RMS.
Best regards
The elliptical coordinates that we use in Calculus are:
x=a*cos(theta)
y=b*sin(theta)
where theta is in [0,2*Pi), to be precise.
Then:
x^2+y^2=a^2*cos(theta)^2+b^2*sin(theta)^2=(a^2-b^2)*cos(theta)^2+b^2
What coordinate system did you have in mind?
Dear Prasenjit Roy
The relation between x2 + y2 and a2 + b2
Following the notations in the attached image:
The Sum x²+y²
In fact as the new axis y = (tanθ)x and y = (tan(θ+(π/2)))x
where θ is the angle of rotation.
The equation of the Ellipse is
(x/a)² + (y/b)² = 1... (E)
(y=(tanθ)x)∩(E) =(m,n)
we get ( x/a)²+((tanθ)x)/b))²=1→x²(1/(a²) +((tanθ)²)/b²) = 1 →
m² = (1/(a²tan²θ+b²))a²b²→
m = (ab)/(√(a²tan²θ+b²))), n=((tanθ)ab)/(√(a²tan²θ+b²))
x²= m2 + n2 = (ab/(√(a²tan²θ+b²))²+((tanθ)ab)/(√(a²tan²θ+b²))² = a²b²((tan²θ+1)/(a²tan²θ+b²))= a²b²(((secθ)²)/(a²tan²θ+b²))
x = ((secθ)ab)/(√(a²tan²θ+b²)))=((ab)/((cosθ)√(a²tan²θ+b²)))
similarly, determine (y = (tan(θ+(π/2)))x)∩(E) = (s, t)
we obtain y = sqrt(s2 + t2 )= (((sec(θ+(π/2)))ab)/(√(a²tan²(θ+(π/2))+b²)))= (ab)/((sinθ)√(a²cot²θ+b²)).
Finally, we calculate
x² + y² = ((ab/(√(a²sin²θ+b²cos²θ))²+( ab/(√(a²cos²θ+b²sin²θ))² = a²b²(a²+b²))/((b²cos²θ+a²sin²θ)(a²cos²θ+b²sin²θ).
Concerning your question:
The answer is given by the following remark:
Remark: x² + y² = a² + b² if and only if θ = kπ or θ = (π/2)+ kπ where
k is any integer and θ is the angle of rotation.
Apply the above calculations to find any term as AM and RMS.
Best regards
Issam Kaddoura Thank you Issam. I was indeed looking for this. But the math was long forgotten for me.
Thank you for reminding.
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