Re: Simplifying the angle between two vectors
- To: mathgroup at smc.vnet.net
- Subject: [mg104172] Re: [mg104122] Simplifying the angle between two vectors
- From: "David Park" <djmpark at comcast.net>
- Date: Wed, 21 Oct 2009 06:33:44 -0400 (EDT)
- References: <26947503.1256029264141.JavaMail.root@n11>
I don't think the angle is generally constant. Only if a == b == 1. Use
VectorAngle. You can write the tangent as: epicikloida[a, b]'[t].
epicikloida[a_, b_][
t_] := {(a + b) Cos[t] - b Cos[(a + b) t/b], (a + b) Sin[t] -
b Sin[(a + b) t/b], 2 Sqrt[a*b + b^2] Cos[a t/2 b]}
VectorAngle[{0, 0, 1}, epicikloida[a, b]'[t]];
ang[a_, b_][t_] =
Simplify[%, a > 0 \[And] b > 0 \[And] t \[Element] Reals]
ArcCos[-((a b^(3/2) Sin[(a b t)/2])/Sqrt[
2 (a + b) - 2 (a + b) Cos[(a t)/b] + a^2 b^3 Sin[(a b t)/2]^2])]
Plot[ang[1, 1][t], {t, 0, 4 \[Pi]}]
Plot[ang[1, 2][t], {t, 0, 4 \[Pi]}]
David Park
djmpark at comcast.net
http://home.comcast.net/~djmpark/
From: Mali_42 [mailto:bloomdalium at gmail.com]
epicikloida[a_, b_][
t_] := {(a + b) Cos[t] - b Cos[(a + b) t/b], (a + b) Sin[t] -
b Sin[(a + b) t/b], 2 Sqrt[a*b + b^2] Cos[a t/2 b]}
depi := D[epicikloida[a, b][t], t]
Arccos[({0, 0, 1}*depi)/(Norm[{0, 0, 1}]*Norm[depi])]
I don't know how to simplfy last line. It needs to be constant(angle
between vector (0,0,1 ) and tangent to epicycloid).
Thanks