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# Distance between two points on a arc

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### #1Cibressus  Members   -  Reputation: 100

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Posted 21 March 2004 - 11:47 AM

how do i figure out the distance between two points on a arc?

### #2Duekiller  Members   -  Reputation: 122

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Posted 21 March 2004 - 01:34 PM

get the point coordinates and use the pythagoran theorum.

### #3Cibressus  Members   -  Reputation: 100

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Posted 21 March 2004 - 01:36 PM

no, thats not quite what i mean, i mean following the arc.

### #4nate0326  Members   -  Reputation: 138

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Posted 21 March 2004 - 01:37 PM

Do you have the points coordinates, or do you have the angle seperation?

[edited by - nate0326 on March 21, 2004 8:37:58 PM]

### #5Nervo  Banned   -  Reputation: 344

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Posted 21 March 2004 - 01:41 PM

well:

the key is knowing the angle between two radii that will subtend an arc. Is that what you wanted to know?

EDIT: Also the angle is measured in radians. Additionally the arc must be a part of a circle. Is the legnth of the arc you are looking something else? If so you will likely need to use a bit of integral calculus.

[edited by - nervo on March 21, 2004 8:45:22 PM]

### #6lezac  Members   -  Reputation: 122

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Posted 21 March 2004 - 01:50 PM

Could anybody tell me how to calculate maximum and minimum helicopter''s weight and how to calculate helicopters gravity center???

### #7 Anonymous Poster_Anonymous Poster_*   Guests   -  Reputation:

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Posted 21 March 2004 - 02:04 PM

If you''re working with cartesian coordinate (x,y,z), get the 2 vectors from the center point to the points of interest (1 vector to each...), from the dot product theorum get the angle subtended by the vectors (in radians), then as Nervo said its just a case of multiplying the radius of the circle by the angle subtended by the vectors ... which you just got from the dot product - that will give you the shortest distance from A to B along the surface of the sphere.

FWIW: the shortest distance between 2 points on a sphere is along the ''great circle'' that both points lie on.

### #8hplus0603  Moderators   -  Reputation: 3293

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Posted 21 March 2004 - 02:35 PM

Hint: on a circle with radius 1, the distance between two points where theta equals pi is, well, pi.

### #9Timkin  Members   -  Reputation: 864

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Posted 21 March 2004 - 03:03 PM

You''re all assuming that the arc is subtended by a radius vector of constant length (i.e., that the arc is a segment of a circle). If this is the case that Cibressus is interested in, then all is well and good. However, if it is NOT the case, then your solutions are not valid (at least, only valid for a circular arc).

In general, you need to integrate the parametric derivative along the arc between the two points. So, you might have an arc defined by a set of parametric equations; e.g., x = x(u), y=y(u) (I''ll assume 2D for the rest of this, but the extension to higher dimensions is straightforward and obvious). Now, by Pythagorus'' theorem, we know that for Cartesian spaces
`ds/du = ((dx/du)2 + (dy/du)2)1/2 `

Therefore, the distance along the arc is given by the integral of ds/du w.r.t u.

If, however, your curve is defined by a function y = f(x), then you have two choices... either transform into parametric form, or utilise the integrand of
`ds/dx = (1 + (dy/dx)2)1/2 `

and integrate with respect to x.

It can be shown that the resulting distance is equivalent to the distance obtained if we integrate the absolute value of the derivative of a radius function from a given point to points on the curve. That is, if the curve can be defined by the vector function r (u), then
`ds/du = |r  ''(u)| `

Any of this can be obtained from an introductory college text on Calculus and Analytic Geometry, which your local library should have.

Good luck,

Timkin

### #10Nervo  Banned   -  Reputation: 344

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Posted 21 March 2004 - 03:33 PM

Timkin, I went with the common assumption that he may have wanted the simpler circular arc, but I also pointed out that calculus may be needed if it wasn''t.

s = ∫ab √(1 + [f''(x)]^2)dx

if you have a curve that you represent in cartesian coordinates f(x), taking the square root of its derivative plus 1 and integrating the result gives you your length between a and b inclusively.

### #11Timkin  Members   -  Reputation: 864

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Posted 22 March 2004 - 01:00 PM

quote:
Original post by Nervo
I went with the common assumption

I generally find that assumption creates more problems than it solves!

Perhaps you should have added the calculus to your original post... that little extra effort would have saved the need for the half dozen other posts that followed!

Cheers,

Timkin

### #12Zipster  Crossbones+   -  Reputation: 398

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Posted 22 March 2004 - 07:44 PM

quote:
I generally find that assumption creates more problems than it solves!

Assume - Makes an ass out of u and me! Hah, I love that one!

Oh, and some related equations to add value to this post - `R = (r2 + a2) / 2a`, but I'll leave it just vague enough so you have to work for it. Has to do with lenses and spherometers. Ok I'm done now.

[edited by - Zipster on March 23, 2004 2:44:54 AM]

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