Re: Spring rates for VP props

From: <mkirda_at_sbcglobal.net>
Date: Fri, 15 Feb 2013 17:51:32 -0000

Hi Nick.

I fully realize that the correct answer might be 'It depends'.

One approach is to graph the torque curve of the rubber batch you use by say every 50 or 100 turns. Measure or guess the RPM average for the first 8 or 10 minutes. At that point you can use the RPMs to get a starting point for where the upper level change could start.
I.E. I get 1200 turns into the motor. I have RPM of 50 and want to start at 8 minutes. So 1200-(8*50)=800 turns... Where is my torque level at 800 turns remaining? Guessing where to have the changeover completed - Maybe 0.07 to 0.06 in-oz? Then design a spring around those numbers.

A spring that changes from 0.12 to 0.06 in-oz will look very different than one that changes from .1 - 0.08.

What works for you?

Regards.
Mike Kirda





--- In Indoor_Construction_at_yahoogroups.com, Nick Ray <lasray@...> wrote:
>
> >
> >
>
> I reread this and realized it wasn't really saying what I wanted it to. The
> more efficient the a model is the wider the range of torque will appear to
> be. i.e. It would look like its taking longer to change over.
>
> The actual range of torques is fixed. Because the spring constant is fixed,
> the force the spring generates to oppose the force of the rubber is given
> by tau = -k (the spring constant) * theta in radians. So, if theta can only
> go between .6 radians and .7 radians because of the high and low pitch
> stops, the force that spring can generate is between .6k and .7k.
>
> I'm not starting a 0 radians because of the preload normally used on VP
> springs.
>
>
>
> > >>"The more efficient the model is the wider the torque range can be."
> > Because with respect to energy usage, everything happens more slowly as the
> > model becomes more efficient.
> >
> >
>
Received on Fri Feb 15 2013 - 09:51:34 CET