stoutfellow: Joker (Default)
[personal profile] stoutfellow
I've put a few final comments on curves under the cut.

1) Earlier, we restricted attention to particles moving along curves with a constant speed of 1. Intuitively, this is always possible on a regular curve. Furthermore, there are only two distinctively different ways of doing this, on any given curve - tracing the curve "forward" and "backward", so to speak. (Actually, there's one other thing you can do, by setting the zero-point at different places along the curve, but that has no effect on the things we've been looking at.) In this sense, then, these quantities - the tangent, principal normal, and binormal vectors and the curvature and torsion - really do describe the curve, rather than any specific way of moving along it. Reversing the direction in which you traverse the curve has the effect of negating the tangent and principal normal vectors and the torsion, but it leaves the binormal vector and the curvature alone.
If you have a regular curve which isn't unit-speed, you can "reparametrize" it to become a unit-speed curve. Pick a zero-time t0. Set s=g(t) to be the integral of |α'| from t0 to t. g(t) is then a one-to-one differentiable function, and its inverse function is also differentiable; that is, t can be regarded as a differentiable function of s. If you then write α as a function of s, it becomes a unit-speed curve. In practice, this is often impracticable; if you remember the problems involved with computing arc length, you'll see why.

2) The Frenet-Serret apparatus can be defined and computed for a regular curve which isn't unit-speed; the formulas are rather complex, though, so I'm not going to say anything about them.

3) For those who remember some of their calculus, I'll toss out this fairly simple exercise. Define α(s)=(3/5 cos s, 3/5 sin s, 4/5 s). This curve is a circular helix. It's not too hard to compute the Frenet-Serret apparatus of this curve; if anyone wants to try it, you can post your results in comments and I'll confirm or deny.

Date: 2006-01-24 03:25 am (UTC)
From: [identity profile] countrycousin.livejournal.com
I've held off a bit, but you don't seem to be getting a lot of homework turned in :<) , so I won't try to hide this.

a = (.6 cos s, .6 sin s, .8 s).
T = a'/a = (-.6 sin s, .6 cos s, .8).
T' = (-.6 cos s, -.6 sin s, 0), so
κ = .6 and
N = (-cos s, -sin s, 0).  
B = T x N
   =
 ( .48 sin s, -.48 cos s , .6)

B' = (.48 cos s, .48 sin s, 0 ) so

τ = .48

Date: 2006-01-24 03:30 am (UTC)
From: [identity profile] countrycousin.livejournal.com
that's T = a'/a' - what I did, not what I wrote :<{

Date: 2006-01-24 10:52 am (UTC)
From: (Anonymous)
Close. In computing B, you crossed T with T' instead of with N, which also threw off your computation of τ.

Date: 2006-01-24 01:36 pm (UTC)
From: [identity profile] countrycousin.livejournal.com
arrgh. You're right, of course. Mostly. It looks like I did the z component correctly, then regressed. Sloppy.

Date: 2006-01-24 11:17 am (UTC)
From: [identity profile] stoutfellow.livejournal.com
Sorry, "Anonymous" was me. For some reason, LJ keeps logging me off.

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