Jamt, put into your simulator an end point just slightly into the halfpipe. Will the oscillations grow? No. Why not? Now put in a curved bottom. Will the oscillations grow? Yes. Then deep in your heart you will understand the mistakes you made about not understanding how parametric oscillation related to swings and halfpipes is entirely about angular momentum, you'll understand that this entire flat bottom discussion is a completely irrelevant and incorrect trolling tactic by you just to try to put me in my place. You'll understand, maybe, but no one else will, so this forum will never know how clueless you really were.
If I retract an inch before the flat plane instead of one inch later would not make significant difference? or 1mm?
I have not said anything against parametric oscillation, except that you don't have to use it because it is just too complicated. Differential equations when you can use a simple energy argument? I suspect also that it was a term used to silent people with less math background.
I made an apology in PM to you stating that part of the reason I made my initial response was that it was a bit of fun to give an example that countered someone who claimed to be on a physics professor level. Obviously that did not go so well and you immediately used that information from our PM conversation in public. You have my permission to publish our entire PM conversation if you want, but not extracted and rephrased parts.
Jamt said a factor of three. But, it will depend on the size of the half pipe. That's why I asked for the radius to be included in that calculation for slope and half pipe size. The greater the radius, the faster you go, the greater the centrifugal force, the greater the energy you can add. When pushing on a curved bottom there is no limit to how much power you can add. You can always create larger forces by making a bigger halfpipe. So, to answer CTkook's question, the size and shape of the moguls are extremely important.
the energy when dropping in to a half pipe is mgh and if we assume a circular shape h=r so the energy is mgr
This is converted to kinetic energy at the bottom and if we for simplicity assume a point mass this energy is
The centripetal force is then:
Fc=mv^2/r = 2mg
So the centripetal force Fc at the bottom is 2mg and then independent of the radius
The total force is Fc+gravity =Fc+mg
That is why I said a factor of three.
If the mass is not a point mass some of the energy will be rotational around the CoM, and thus the force is a bit smaller than this and depending on the moment of inertia of the body.
I don't know the moment of inertia of a crouched skier but I suspect it is significantly smaller so that Ftotal=3mg is still a reasonable approximation. Maybe I'm wrong about that approximation, I don't know but you are welcome to check if you know.