Calculate the arc length between two points over a hypersphereby 7toni7 Tags: arclength, distance, geodesic, hypersphere 

#1
Dec1312, 04:22 AM

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Good morning,
I'm trying to compute the arclength (geodesic distance) between two ndimensional points over a ndimensional 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, 



#2
Dec1312, 05:21 PM

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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 dotproduct) 



#3
Dec1412, 04:48 AM

P: 7

Hello tinytim,
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. 



#4
Dec1412, 05:23 AM

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Calculate the arc length between two points over a hypersphere
hello 7toni7!
2. yes: (a_{1},a_{2},…a_{n}).(b_{1},b_{2},…b_{n}) = a_{1}b_{1} + a_{2}b_{2} + …a_{n}b_{n} (don't forget that the dot product gives you R^{2}cosω, so you'll have to divide by R^{2}, and then use the cos^{1} button ! ) 3. yes 



#5
Dec1412, 07:03 AM

P: 7

Thank you.
Then, the arclength on a nsphere can be computed as follows: S = R*acos(a.b/R^{2}). I think it is correct. Isn't it? A last question, do you know how to compute the intersection point between a nvector and a nsphere? Thank you so much again. Best 



#6
Dec1412, 07:58 AM

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do you mean an nvector starting from the origin (the centre of the nsphere)? if not, how are you defining the nvector and the nsphere? 



#7
Dec1412, 08:44 AM

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Hello,
Yes, suppose that we have one nsphere. Inside it, we have a npoint (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 nsphere passing from H) with the nsphere. do you understand? is it possible? Thank you in advance again, Best. 



#8
Dec1412, 08:52 AM

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then yes, it's easy … the nvector to the intersection will be a scalar multiple of the nvector to H, such that the magnitude of the nvector (ie, the squareroot of the dotproduct with itself) equals the radius 



#9
Dec1412, 09:39 AM

P: 7

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 



#10
Dec1412, 11:02 AM

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Q = R*(P/P) 



#11
Dec1412, 11:11 AM

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In ndimensional 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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