Physics Question

Mark Chisholm

Mark Chisholm

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Here is a portion of an email that I received from a friend yesterday:

I wonder if any of you have a more exacting method to predict shock load forces. Don Blair’s rule for shock loading - which is based on Newtonian physics - is that for every foot an object falls, it gains a unit of its weight plus one. I would apply Newton’s own formulas, but I’ll admit that I was so bad in physics that I was forced to take it in summer school.

Does anyone have any good info that I can pass along? He's working on an article.
 
I don't think there is a real "rule of thumb". Here's what TK and I have found out during some demo's using a Dillon digital dynamometer. We drop a 15 lb. log 2 feet and the dyno reads anywhere from 100 to 105 lbs. That's 6 to 7X. It will depend on rope construction, how much rope is "in the system", how much "stretch" is taken out before the drop, how much the tree absorbs. Alot of variables to consider. This is why we work on a 10 to 1 safety factor with software. Should it be more???
 
To calculate impact force (known as impulse)

I=FxT (impulse=force x time) measured in KN on most equipment

The easy part of the equation is Force, the near impossible part is Time. Time would be measured from when the rope goes taught, through all the stretching and deflection, until the system goes static or begins rebound. We can all imagine that this time would be measured in units less than seconds. Probably the only way to measure it would be with a high speed camera or some pretty rare electronic sensors.

A stopwatch would be useless because the difference between .4 and .2 seconds gives double the impulse.

Each small variable change, stretch in rope, length of rope, friction in block, slip of friction device, weight of object and deflection of tree (which would be measurable to the roots) would affect the impulse.

No one has done this yet because there are to many variables to do it accurately. You could measure everything to get a result but when you set it up somewhere else the results will be very different.

Dave

/forum/images/graemlins/drowninga.gif /forum/images/graemlins/drowninga.gif
 
I think the easiest way to explain it is to work it out in seconds of freefall
A body accelarates at 10 metres per second squared well it is actually 9.8 but we will say 10.So we would say that after 1 second of freefall a body would be travelling at 10m per second so if you times the weight of the object by 10 you get the energy that is going in the rope after 2 seconds its travelling at 20m per second
Times the speed by the weight and you have you answer....... and so on

Thats about as simple as I can make it.
Fall factor 2 static drop is always going to be around 10m per second Unless you let it freefall along way,but most of us try to get the block as close to the system as possible to avoid this.So generally we would never have a body in freefall for more than a second and of course we don't drop static if we can help it.....


Didj
 
Thats a nice one Dave but its like you said before to work it out you need all the variables and even with this conversion programme you aren't ever going to be accurate even with the weight,we can all make a guess at the weight to be rigged but we are not ever going to know exactly unless we weigh it after its been dropped and then its to late.I think its always more about knowing the potential for increased forces than getting exact numbers.That way we should always over compensate for it.

Cheers Didj
 
[ QUOTE ]
To calculate impact force (known as impulse) I=FxT (impulse=force x time)

[/ QUOTE ]
A simple way to understand shock loads is this:

Since the kinetic energy of a falling object is its weight times the distance of fall, the impulse can also be measured by the ratio between distance fallen (acceleration) and the distance of stopping (deceleration).

If an object falls 10' and is stopped in 2 feet of stretch or dynamic belay, then the average impulse force (or change of momentum) is 10/2 = 5 times the weight.

While this simple formula won't give the peak impulse, the average is probably a close enough approximation for a rule of thumb.

And, while the time of deceleration may be impossible to calculate in the field, the distance of deceleration is more obvious.

- Robert
 
Will this help?
5300184.jpg


or this?
liv_life_preservers.jpg


or maybe this?
Life Preserver for entering a Black Hole
 
If that's true, he looks a little like Liv Tyler. Just goes to show you can't always judge a book by its cover.

I don't mind admiring a man for his good looks; I just prefer to know he's a man so I can keep the admiration in the proper perspective.
 
[ QUOTE ]
Glen, that's a transvestite.

[/ QUOTE ]

Hey glen kevin found that out after he pulled the life preserver from under the seat. /forum/images/graemlins/pokinit.gif
 
***Shock Loading***

Spring Energy = 0.5kx^2
where k is the spring constant (F=kx, Hooke's Law) in force/displacement units
x is displacement

Potential Energy = mgh
where m is mass, g is accel due to gravity, and h is elevation change

Now, assume that all the potential energy is converted to spring energy, therefore set PE=Spring Energy

mgh=0.5kx^2

First, you have to know how much a rope stretches at a given force, and use it for k, assuming a fairly linear relationship. k is easy to figure out by measurement.

Now, you can solve the last eqation for x to figure out how much stretch your Potential Energy will cause. Plug x into F=kx, and you've found force. DONE! :)

DO NOT TRY TO MEASURE IMPACT TIME! It won't work. Use the above energy approach.

Cheers,
roo
http://www.geocities.com/roo_two/knotindex.html
 
Excellent Knot sight Bro!

Have had that link and passed it on for quite some time; is also listed on my sight a few times; great to see ya here!




If you hang a pulley as support and anchor 1 end, and hang 400# from the other, the support carries 800#. If you use a DWT from a single point/from dropeye pulley, the same 400# only places 1.5x Load on support; or 600#. So that is better, plus, you can pretension the system more too. Beginning line tension is somethiong i see missing from force calculators, that i beleive Isaac was addressing.

The DWT, as far as the line goes, could hold more weight than a single line too. :: The same materials and constrauction of line is now at higher tensile/ greater SWL to the 400# Load. ::Less elasticicty/dynamic support; so on real dynamic force/movement less 'support' dynamically than a single line.

The DWT that gave less Support Load in a static hang, now gives more Support Loading than a single line, when the elasticity quotient matters. So, to make that the wiser selection i think we need to maximize the added pretensioning 'gift', while minimizing the dynamic loss 'tax' from the decision choice of DWT over single line. (Find the forces that stand fer ya, and thsoe that stand agin' ya!). So pretighten more, for less dynamic shock; use where the dynamic loading doesn't go very high; or a single line would be better for a weaker support to get less dynamic loading on it.

Even though, at a static hang the DWT is less Support Load; in real dynamic support situations, the DWT loads Support More. The 'tax' then sets agianst ya in the life of the line, measuring line life how mountain folk do, by using up the elasticity, degrading it, rather than watching for degrading the strength. So, it depends on which you are trying to save more(line or support), in consideration to each's reaction to the loading ranges; as well as the type of loading dynamic or static; which strategy is best.

This is what i meant by trying to show in Rigging Program Thread; as one of first theories proven when i bought it. Though honest, i was kinda trolling in the drama of the procalmation in the sub thread "Weaker tensile can give safer loading". But, it is true, it can sometimes.

Usually adding more line to the system gives more elasticity; the exception being, when that added line length is in legs of support to load. In low elasticity type or length, i think the single line thru pulley would still be a doubling factor, but less dynamic force deleiverd to that doubling. Still 1 1/2x Load in DWT, but almost double loaded going into that multiplier, because of the loass of elasticity percentage.
 
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