Easing into gravity

[Andy24]

I think that last post is a real conversation stopper. I haven't read it yet. so it might have useful information I ddon't know about. Easing into gravity wouldn't help suspention bridges or cable stay bridges. although it would relieve them of a lot of stress when they settle. mabey we want easing into gravity as an add on for the next version and chronic logic should put pre tensioning into the next game.

(Edited by Andy24 at 4:11 pm on Nov. 4, 2001)

[mendel]

I got completely confused reading VRBones post of Nov 6, 6:01pm (6:25 edit) because I could not make heads or tails of the various additional forces, steps, cycles, and waves. However, I managed to think about it on my own, and even uncovered another error that ocurred to me in my talk with baggio.

I found that the final energy dissipated by the system does not stay equal with any gravity ramp (as I erroneously stated), but potentially decreases as the number of steps is increased.

I found this out by first looking at a vertical bar fixed at one end, without damping applied; if you go to 0.5 g and then take it to 1g as the oscillation is at its lower extremity, it will come to rest immediately; if you do that step at the upper end of the oscillation, you might as well not have bothered with 0.5 g.

Also note that at static equilibrium point, internal energy is equal to kinetic energy; it is half the work done by gravity (m*g*y, y being the displacement downwards) because internal beam force starts at 0 and is m*g only at equilibrium.

Now think of that same bar; at final g, let it's gravitational force be G, with an equilibrium displacement of y. Ramping up g in n steps and assuming we let the structure reach equilibrium between the steps gives us the work Wi per step i as displacement per step (y/n) * strength of gravity force at that step (i/n)*G :
Wi = (y/n) * (i/n)*G
Summing up Wi for i=1 to n gives us total work done:
W = G*y/2 + G*y/(2*n)
The first term G*y/2 is equal to the internal beam energy at equilibrium; the second term is equal to the excess energy that must be damped away. This energy decreases in inverse proportion to n.

However, this result is at best an average value when the system is not at rest between the steps.

What we do not know is how damping is affected by this; I still fear that such a procedure might increase the time the structure needs to settle, and the right timing for the steps has not been determined, either (and might vary with the structure, so might have to be dynamically computed).

In hindsight, VRBones first paragraph now makes some sort of sense to me; paragraphs 2 and 3 still leave me stranded.

[Andy24]

I was thinking about easing into gravity and I thought of something that I havn't seen mentioned here.

I was thinking about the bridges that are designed to fall into place. wouldn't it make them unrealisticly strong.

An alternative idea: Make all links *unbreakable* until the bridge has settled. This would ensure they would only have to withstand the stresses caused by bridge use, and not those initial settling forces.

(Edited by Chillum at 7:11 pm on Nov. 7, 2001)

[VRBones]

Quote: from mendel on 8:31 am on Nov. 7, 2001
I got completely confused reading VRBones post of Nov 6, 6:01pm (6:25 edit) because I could not make heads or tails of the various additional forces, steps, cycles, and waves.
Sorry, was trying to put it in more layman's terms than pure math, I'll try to flesh it out a little. Made perfect sense to me at the time ;)

I found this out by first looking at a vertical bar fixed at one end, without damping applied; if you go to 0.5 g and then take it to 1g as the oscillation is at its lower extremity, it will come to rest immediately
My second and third paragraphs are do do with this effect. Applying the second increase at the lower extremity of the oscillation cancels out each other due to the waves of oscillation of equal amplitude being 90 degrees apart. The same can be said if you apply the additional force in 4 stages of 0.25 G at 45 degree increments. Extending this you can see that if an equal amount of force is applied each step, and that the steps are applied evenly across the whole oscillation, they will cancel each other out. Although we don't know the frequency of the oscillation of each bridge, if we assume that all steps apply the same force evenly and that the time to apply these steps is greater than the minimum frequency of any bridge, this effect will come into play.  

This is all without damping though. When you introduce damping, the remaining force from the initial step that is still oscillating in the system at a point in the future is proportional to the damping co-efficient (another way of stating the definition of the damping co-efficient). Since the amount of additional force is equal at each step, the net resulting force is equal to the amount of force damped over half an oscillation. The higher the damping effect, the higher the net resulting force.

If you managed to work out the damping co-efficient of a bridge, you could also proportionally reduce the force at each step by that amount, therefore the resulting force from half an oscillation ago would equal the new force being added, thus returning the equation back into equilibrium. However, from some quick tests the damping co-efficient seems to vary across bridge designs (proportional to the amount of regidity across the whole bridge ?).

So from all that, what's the conclusion? If you had the amount of force applied each step slowly decreasing, and the time taken to apply the whole gravity greater than one oscillation, you would get a pretty stable result. The smaller the steps the better, but means the time to apply all the gravity is more, approaching infinity (and no-one is going to wait that long ;) ). I think around 3-4 oscillations worth would be reasonable, so assuming an oscillation is one second (I've seen oscillations of .5 seconds and as long as 2 secons) and that 50% or the resulting force was damped over that time, you would have the bridge gradually sink due to the strain for 4 seconds, then have a remaining oscillation of ~4% the size of the oscillation experienced now.  

(Edited by VRBones at 4:26 pm on Nov. 7, 2001)

(Edited by VRBones at 4:27 pm on Nov. 7, 2001)

[Andy24]

although this may not work quite as well it would help large bridges from bouncing for 30 seconds. and this would be much easier to implement.

[falkon2]

Isn't this being discussed!?