Vector analysis / dynamics Math question

Hi Brad,

If you don't get answer here I recommend to ask in one of the sci.math.* or alt.sci.math.* groups.

Greetings from Germany Joerg

most introductory dynamics texts use a similar setup to prove calculus can be used (as del x goes to dx) to solve these type problems. My kids book from the U of M uses the same approach - but I don't think they use rotation - the proofs i have seen used vectors, so i am not sure you will find a modern one using rotations.

If i have it right (and I may not cause I got in late and I have to get moving early - )

1)those rho terms are the radii: rho in the xz plane and rho primed in the yz plane - that leaves the radii of the xy plane to be derived. 2) the displacements are p, q, and r - straightforward, since motion is only in the x direction

so then from arc length = radii x radians, that gives radians of rotation = arc l/radii = dx/rho

that means that the radius of angle phi3, which is a function of dx and dy because it is the rotaion in the xy plane, is the derivative of the xz plane's y component's.

drat-- sorry - gotta send NOW - out of battery - more later in the week .

sorry bout that - charger is in the car,

it is zero where dx is zero

Hi Hob,

Yes, I agree, p, q and r are straightforward.

So, rho x dphi2 = -dx follows easily from what you say above.

I have agonized over the previous paragraph, but I haven't grasped the meaning yet.

One thing I am looking at is to let k=1/rho and k'=1/rho'

so k and k' are the curvature of the sections xz and yz

the equation then becomes (k-k') dphi3 = d(k)/dy dx

An explanation regarding k I obtained from a textbook ...

If the angle between the tangents to a curve at r and r + delta r is delta theta, curvature, k = dtheta/ds

In other words, k is the rate of change of the angle of the tangent to the curve as arc length changes.

This led me to write:

(dtheta/dx - dtheta'/dy) dphi3 = d(dtheta/dx)/dy dx

I fudged ds to be approx. dx and dy on the respective sections xz and yz

Then the wheels fall off, can't seem to get anywhere with this approach. Any help much appreciated.

Cheers, Brad

back to this after a short hiatus ---

so we have the linear displacements in the x plane, y plane, and z plane as p, q, and r

the rotational displacements, in radians, are derived from the curvature radii and the displacements in the three planes, using the arc length-radii-angle equation. e.g.,:

rho in the xz plane as a function of x,z, ; rho primed in the yz plane as a function of y,z ; and the rotation in the xy plane, which is derived from the curvature in the other two planes, i.e., a function of rho and rho primed with regard to x.

(Since there is no z component in the xy plane, it does no good to differentiate in that plane with respect to z --- so now, the resultant equation for angular movement in the xy plane will then be of the form dx, dy.)

then -

1) the angle = displacement/radii

and here's where I am going by feel --

to get the partial differential both in x, but also having to remove the dy component of the other rotations which gives the z rotational displacement in the plane, you have to divide out dy

so

2) phi3 = [ d(function of the other two plasnes only in x,y) * d(x)]/[d(y) * rho]

or, the third axis angular rotation, that in the xy plane, is

3) phi3 = [d[1/rho]dy]* dx

that's my best guess -