I completely agree with pdql. Im simple minded and this is the way I see it. If what you are saying were true, I would be better off jumping off a bridge attached to 1" manila rope instead of a bungee cord.
I completely agree with pdql. Im simple minded and this is the way I see it. If what you are saying were true, I would be better off jumping off a bridge attached to 1" manila rope instead of a bungee cord.
Again, that just isn't true.
Consider a different example of similar physical properties: You are in a car, doing 60 mph. You slam on the brakes and stop in 100 feet, or...same speed, same car, but you hit the end of a huge tow strap that stretches down to a stop in 100 feet. You will naturally be snatched backwards at the end of tow strap "stop".
Which method of stopping has the greatest acceleration force? You just went from 60mph to zero in 100ft. The deceleration is essentially the same, although braking tends to stop fastest in the beginning and then fade, whereas the tow strap will stop with the greatest rate at the end.
As I hope you understand, the stopping forces are not different, they just dissipate the energy differently. When the brakes stop the car, the tires burn rubber, the brakes heat up, and the suspension bounces down in the front. Energy is converted to heat. This is what happens to your "run to a stop" log scenario.
When the tow strap stops the car, some of the energy is converted to heating the strap, but most of the energy is used to stretch the strap. When all the energy of the car is spent, the strap yanks the car backwards because the strap has stored the momentum of the car, rather than dissipating it as heat.
The bouncing log acceleration is really two different acceleration calculations. The first occurs when the log is slowing down, and the second occurs when the rope yanks it back. Since the rope cannot possibly store more energy than the initial fall, the second acceleration is smaller than the first.
If you still don't see that bouncing backwards does not double the acceleration force, imagine that the tow strap ruptures, right as the car comes to a stop.
I have been putting together a rudimentary forces calculator for my own use as a trainer I would like someone to check my formulas, and suggest other variables that I may have overlooked, and as I'm sure you would do anyway critique it in general.
I am aware of the many limitations and inaccuracies of a calculator... this is only for training and expressing the importance of letting the rope run etc.
the variables that are taken into consideration are:
mass of log
dropping distance (x the force of gravity)
rope length
rope stretch
branch flex (movement at the high point)
run in the rigging line
the formulas are:
force = mass x acceleration
impact force = mass x acceleration / breaking distance (assuming breaking is even)
so for example a 300kg log dropping 1m and the breaking distance is the sum of limb flex 10cm and rope stretch 45cm (15m rope in the system x 3%rope strech) and the run the log is given through friction device
so if the log isn't let run the impact force would be
300kg x 1m x 9.8/(.1+.45)= 5349N
and at the block it would be 5349x1.66=8879N
if the log is let run even 50cm the impact force would be
300kg x 1m x 9.8/(.1+.45+.5)= 2802N
and at the block it would be 2802x1.66=4651N
any feedback would be appreciated.
thanks.
Dan
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