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Calculate the arc length between two points over a hyper-sphere |
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| Dec13-12, 04:22 AM | #1 |
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Calculate the arc length between two points over a hyper-sphere
Good morning,
I'm trying to compute the arclength (geodesic distance) between two n-dimensional points over a n-dimensional sphere (hypersphere). Do you know if it is possible? If yes, please, I'd be very pleased if you, as experts, provide me this knowledge. Thank you very much Best, |
| Dec13-12, 05:21 PM | #2 |
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hi 7toni7! welcome to pf!
![]() won't it just be the radius times the angle between them? (which you can get from the dot-product) |
| Dec14-12, 04:48 AM | #3 |
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Hello tiny-tim,
Thank you very much for your answer, and I'm pleased to be in this forum. Yes, I think the same. In 2D and 3D is just: (arclength = S, radius = R (in radians), angle between points= ω) S = R*ω. Then, I have 3 doubts: 1 - when dealing with dimensions greater than 3...the order of hundreds, could we do the same computation than 2 and 3 Dimensions? Could we extend this equation to higher dimensions? 2 - The dot product in N dimensions...is just the same than 2 and 3 dimensions? 3 - This formula is in an euclidean space, isn't it? Thank you very much, Best regards. |
| Dec14-12, 05:23 AM | #4 |
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Calculate the arc length between two points over a hyper-sphere
hello 7toni7!
![]() 2. yes: (a1,a2,…an).(b1,b2,…bn) = a1b1 + a2b2 + …anbn (don't forget that the dot product gives you R2cosω, so you'll have to divide by R2, and then use the cos-1 button ! )3. yes
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| Dec14-12, 07:03 AM | #5 |
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Thank you.
Then, the arclength on a n-sphere can be computed as follows: S = R*acos(a.b/R2). I think it is correct. Isn't it? A last question, do you know how to compute the intersection point between a n-vector and a n-sphere? Thank you so much again. Best |
| Dec14-12, 07:58 AM | #6 |
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![]() do you mean an n-vector starting from the origin (the centre of the n-sphere)? if not, how are you defining the n-vector and the n-sphere?
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| Dec14-12, 08:44 AM | #7 |
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Hello,
Yes, suppose that we have one n-sphere. Inside it, we have a n-point (this point different of the origin, it is another point named H). So, I have to compute the intersection of the line (that goes from the origin of the n-sphere passing from H) with the n-sphere. do you understand? is it possible? Thank you in advance again, Best. |
| Dec14-12, 08:52 AM | #8 |
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then yes, it's easy … the n-vector to the intersection will be a scalar multiple of the n-vector to H, such that the magnitude of the n-vector (ie, the square-root of the dot-product with itself) equals the radius
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| Dec14-12, 09:39 AM | #9 |
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Well,
This is how I do it in 2 dimensions. See image. Now, my question is: could this development be extended to N dimensions? ![]() Thank you |
| Dec14-12, 11:02 AM | #10 |
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Q = R*(P/|P|) |
| Dec14-12, 11:11 AM | #11 |
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In n-dimensional Euclidean space, the (hyper)sphere with radius R and center at [itex](a_1, a_2, ..., a_n)[/itex] has equation [itex](x_1- a_1)^2+ (x_2- a_2)^2+\cdot\cdot\cdot+ (x_n- a_n)^2= R^2[/itex]. The line through the origin and point [itex](b_1, b_2, ..., b_n)[/itex] is given by the parametric equations [itex]x_1= b_1t[/itex], [itex]x_2= b_2t[/itex], ..., [itex]x_n= b_nt[/itex]. Replacing [itex]x_1[/itex], etc. in the equation of the sphere with those gives a single quadratic equation for t. Finding the two solutions to that equation gives the two points at which the line crosses the sphere.
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| arclength, distance, geodesic, hypersphere |
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