# I What's the equivalent distance of fall for 1500G

1. Apr 8, 2016

### David Wright

Hi friendly physics gurus.

I'm trying to find equations to help answer a question I have inside my own head. I know the weight of an object (w=170g). I know the object will be subject to a shock of (1500G and time to stop t=0.5ms). My question is: what's that like in terms of how high in the air it falls? What I would like to learn is how I can come up with equations to determine the equivalent 'drop-height' this 1500G @ 0.5ms test will simulate. So far, I only understand that 1G = 9.81 m/s². I get lost when I try to use an acceleration equation d = 1/2 at² because I understand that equation to answer the amount of distance it took to stop the object from the moment of impact (initial velocity at the moment of impact) to final velocity of 0. But I'm trying to calculate how high the object was before it was dropped to experience that impact.

(I think I'm getting confused on initial/final velocities and times and how they relate. When I think about how I am confused about those variables, I get even more confused.)

Any takers? Are there other variables I'm unaware of at this time that could be useful?

2. Apr 8, 2016

3. Apr 8, 2016

### David Wright

Thanks SophiaSimon. That thread helps shed some light on things. However, the answer to my question still eludes me. Somehow there is some time built into the explanation found on that thread hidden between the terms used: like, "rug" and "cement". I can see how a rug provides a longer distance to stop an object than does cement which ultimately increases the amount of time it takes to stop - if I assume the velocity at the moment of impact is the same on both surfaces. I can make some sense of that part.

Do they mean in that thread that to figure out what the initial "drop-height" of an object that stops in "t" seconds from the moment of impact is simply (d = 1/2 at² * G) [bold letter for emphasis on what I understand should be added] where 'd' is drop-height; 'a' is the acceleration to stop (G * gravity); 'G' is the Shock force; and 't' is the time to stop from the moment of impact?

I continue to scratch my chin while looking at that because with my application 0.5ms takes 1.8mm to stop. (1/2*[1500*9.8]*0.5²). However, when I substitute 0.5ms for 1ms as described in the thread you shared, I get 7.3mm to stop. This is where I get confused. I'm trying to translate my t=0.5ms to their 'rug' or 'cement'. According to the thread, 1mm is the stopping distance of carpet, but I'm getting 7.3mm to be the stopping distance.

Eash ... I'm so confused. :(

4. Apr 8, 2016

### SophiaSimon

OK, I think I see your problem.

You know the acceleration the object can withstand is 1500G. You also know the object will go from some initial velocity to zero in the time $t=0.5ms$. You can relate the acceleration of the object to the initial and final velocities and the stopping time of the object using the equation for constant acceleration. The initial velocity right before it hits the ground will depend on the height it falls from.

Assume the object falls from an initial height $h$, and has an acceleration due to gravity. The object will have an initial velocity of zero and some final velocity right before it hits the ground that depends on the height $h$. You can find this velocity using kinematic equations. This is the initial velocity the object will have as it hits the ground and begins to decelerate to a velocity of zero in the stopping time $t=0.5ms$.

Last edited: Apr 8, 2016
5. Apr 8, 2016

### nasu

You cannot establish this equivalence as you want it.
The acceleration of the falling body at the contact with the ground depends on the type of contact, on the surface it falls on.
One thing is to fall on hard concrete and a completely different one is to fall on a trampoline, even if the height is the same.
If you multiply that acceleration by the time you will get a velocity. You can calculate from what height should something fall to gain that velocity. But this will not guarantee that the actual acceleration suffered when stopped by the ground will be the same. The time to stop will depend on the stopping surface.

6. Apr 9, 2016

### SophiaSimon

I see, so we can't just assume the acceleration will be constant as it's stopping, because it depends on the stopping surface. But if we are given a stopping time, which is fairly small, would it not be a good approximation to assume this?

7. Apr 9, 2016

### CWatters

You can't compare a "fall from height" with an "impact shock" for the reasons others have given (not least because it depends how hard the surface is).

However I believe you have enough info to calculate approximately how far the weight did fall if the shock you describe was caused by it being dropped......

You have to make some assumptions that might not be true.. Lets assume that when it hits the ground it experiences constant deceleration = 1500g = 14715 m/s/s and it stops in time t=0.5ms (eg it doesn't bounce). We also need to assume that the weight isn't deformed, only the ground is.

Then you can work out the velocity u that it had when it first contacts the ground using the SUVAT equation..

v = u + at
where
u is the initial velocity
v is the final velocity (=0 because it stops)
a = -14715m/s/s
t = 0.5*10-3

since v=0 we can rearrange to give

u = -at
= 14715 * 0.5*10-3
= 7.4m/s

If the object was dropped from height s we can use another SUVAT equation to calculate the height s...

v2 = u2 + 2as

where
v is the final velocity as it hits the ground = 7.4m/s from above
u = 0 (it was dropped not thrown down)
a = acceleration (in this case a = g = 9.81 m/s/s)
s = height

rearrange to give
s = v2/2g
= 7.42/(2*9.81)
= 0.6m

So to recap our model suggests the weight could have been dropped from around 0.6m and hit the ground at 7.4m/s where upon it experienced constant deceleration of 1500g and stopped in 0.5mS.

We can also estimate how deep a hole it made in the ground (eg the stopping distance)

v2 = u2 + 2as

where now
v = 0
u = 7.4m/s
a = -14715m/s/s
s = depth of crater/stopping distance

rearrange
s = u2/2a
= 7.42/2*14715
= 2mm

That suggests the surface isn't rock hard like steel/concrete nor soft like deep pile carpet. Perhaps something like a vinyl floor?

8. Apr 11, 2016

### David Wright

Excellent reply. I had been too focused on the time element and glossed over v² = u² + 2as. That was the missing piece - and I'm near certain I understand. However, just to be sure I understand, can I confirm the numbers in your calculation? In the examples set above, you arrive at 0.6m to be the height, but I get 3m when I calculate 7.4²/(2*9.81). I'm pretty sure it's just a mis-click or typo, but I've been wrong too many times in my life and thought I would double check if I'm doing the operations of that equation correctly.

9. Apr 11, 2016

### CWatters

Yes looks like I miscalculated.

7.4²/(2*9.81) = 2.8m not 0.6m

10. Apr 11, 2016

### David Wright

This smells like I need to use integrals. So I looked up some more info and scratched my chin until it was a little sore. So far I'm gathering from your comment that the shock impulse should look like the following graph, (that I'm assuming too much that the acceleration is constant) and that I need to invite some integrals to this dance party. Did I read your comment correctly?

11. Apr 11, 2016

### David Wright

Cool. Now my imagination will take these principles and use them with a variable deceleration, rather than a constant one. Thanks!

12. Apr 11, 2016

### nasu

Well, the force increases to a maximum and then decreases back to zero. That much you can assume.
But it does not have to be symmetric around the maximum. And does not have to be parabolic (I suppose your curve is a parabola. At least it looks like one :) ).
What I was trying to say is that the shape and width of this curve depends on the two materials colliding and maybe on the speed of impact.
But still is not too clear what are you trying to do. are these values (0.5 ms and 1500 g) results of some measurement? If yes, was this done on an actual falling body?

13. Apr 12, 2016

### David Wright

Hi Nasu, I think I follow your explanation. It makes sense to me when I exaggerate the surface properties like you did (a trampoline or concrete). In regards to your questions:

1.) The parabola was a generic approximation for the sake of simplicity and also for saving the last of my brain cells for that day.
2.) 0.5 ms and 1500 g are input variables to a test for an object conducted by a third party.
3.) I get the results of this test passed back to me in the terms of 'pass/fail' depending on if the object breaks or not. Without knowing their test procedure, I'm ultimately trying to quantify what the test means in real life - as one who usually sits in the back of a room full of smart people, I have a hard time trying to keep up with those who throw around these kinds of numbers at whim and can comprehend their significance. This is my effort to make more wrinkles in my brain. It's one thing for me to hear "wow, it survived 18 impulses of 1500G @ 0.5ms, eh?", it's another for me to hear "wow, that clumsy guy dropped it 17 times while he crossed the street and then it fell from his hands one last time where he accidentally kicked it to the curb 3 feet away and it still works, eh?"

14. Apr 13, 2016

### nasu

If this is the case, maybe you can ask them what do they mean by that value of acceleration. Is it the average value or maybe the peak value (or something else)?
I understand your goal. Unfortunately there is no a straightforward answer unless you know detail properties of the object itself. As I understand it, the 0.5 ms does not refer to the time the object was actually stopped from free fall but to the time they conducted their (unknown) test.