When I start swinging on a rope I calculate the current kinetic energy
and then use that to determine the angular velocity:
curKineticEnergy = origKineticEnergy + (origPotentialEnergy -
curPotentialEnergy);
curV = Math.sqrt( 2 * curKineticEnergy );
angVel = curV / radius;

If I'm moving only horizontally (say I'm running along the ground) and
then grab a rope that's hanging vertically, then all my velocity is
converted into kinetic energy for rope swinging.

However, if I'm moving vertically (say I'm falling straight down) and
then I grab a rope that's hanging vertically, then all my velocity is
eliminated (because the rope does not stretch and there's no
horizontal movement), so none of the velocity is converted into
kinetic energy for swinging on the rope.

Unfortunately, I cannot figure out a general equation that would allow
me to take my initial velocity vector (origVx, origVy) and subtract
out the parts of that vector that the rope removes, and then translate
the rest into the kinetic energy equation that allows me to determine
how fast the rope should swing.

This is for my game where I'm swinging on a rope. Feel free to take a
look at my previous question to see why I'm asking this.

So, given:
  (origVx,origVy) = original velocity vector
  f = angle from vertical
  (0,0) = point that the rope is attached to
  (curX,curY) = the point we start swinging from

I need to get:
  (resultingOrigVx,resultingOrigVy) = the original velocity vector
that we get after removing the parts that the rope cancels out.

Hello again mxn match

Thanks for the rating and tip on the previous question. 
I was away from Internet for few days, so I just noticed your
new question today.

So, you need the total energy, which is

TotalEnergy= origKineticEnergy + origPotentialEnergy
and also
TotalEnergy= currKineticEnergy + currPotentialEnergy

as it does not change during the swinging...

You need to decompose your 'incomming velocity vector' into two
components, 1) tangential (which will contribute to swinging)
            2) radial (which gets absorbed by the rope

So relative angle  f - a is essential, where a is angle of approach.
When f=a, all is absorbed, and TotalEnergy is just m * g *h ..
Since it (decomposition) is a simple rotation,here, the formula 
for tangential velocity must be someting like  Vapp * sin (f - a) = Vtgn

where Vapp is magnitude of that approach vector, also called speed of
approach.

So your totalEnergy =m * g * h * .5 * m * Vtgn * Vtgn

This I hope is sign independent :-) 

hedgie

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Clarification of Answer by hedgie-ga on 04 Oct 2003 00:35 PDT

errata
 totalEnergy =m * g * h * .5 * m * Vtgn * Vtgn

should be + of course

totalEnergy=origPotentialEnergy + .5 * m * Vtgn * Vtgn

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Request for Answer Clarification by mxnmatch-ga on 05 Oct 2003 18:51 PDT

That works great! Thanks!

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Clarification of Answer by hedgie-ga on 06 Oct 2003 22:29 PDT

Thanks  mxnmatch

 I have to pay more attention to signs (and spelling :-)  )

Assuming the rope is vertical, just ignore origVy (the vertical part
of the velocity vector), and use origVx in your calculations.