Testing the Impact of Different Prototypes: Dropping 20lbs from 10’

  • Thread starter Thread starter bp1945
  • Start date Start date
  • Tags Tags
    Impact Testing
AI Thread Summary
The discussion focuses on testing the impact of different prototypes by dropping a 20-pound weight from a height of 10 feet to assess damage. A formula for calculating the impact velocity is provided: v = √(2ad), where 'a' is the acceleration due to gravity and 'd' is the drop height. The impact velocity at 10 feet is calculated to be approximately 25.36 feet per second. The user expresses difficulty with the math and requests a simplified explanation, which is provided. The conversation emphasizes the importance of understanding the impact dynamics for product development.
bp1945
Messages
3
Reaction score
0
I am developing a new product. I am currently constructing various prototypes of the product changing only one component using a different material with each model.

I want to test the impact results of each different model dropping a consistent weight (i.e. 20lbs) from a consistent height (i.e. 10’) for each test so video these results to determine which model will best suit my needs.

So, this sounds like a simple question, dropping 20lbs from 20’, what will be the velocity at impact? And is there a simple formula I can use if I change the variables?
 
Physics news on Phys.org
bp1945 said:
I am developing a new product. I am currently constructing various prototypes of the product changing only one component using a different material with each model.

I want to test the impact results of each different model dropping a consistent weight (i.e. 20lbs) from a consistent height (i.e. 10’) for each test so video these results to determine which model will best suit my needs.

So, this sounds like a simple question, dropping 20lbs from 20’, what will be the velocity at impact? And is there a simple formula I can use if I change the variables?
If the products that you are dropping are relatively aerodynamic you can use a very simple formula:
v=\sqrt{2ad},
where
v is the velocity when the object hits the ground (ft/s)
a is the acceleration due to gravity (32.17 ft/s2)
d is the height you dropped it from (ft).
 
Hey gamesguru, thanks for your response. After reading your response and re-reading my question I realize I didn't make it quit clear.

I'm not dropping the product, the product will be stationary and parallel to the ground and I will be dropping the weight from a certain height straight down on the product to determine velocity and record (video) impact damage.

And gamesguru, I am worse than terrible at math so that formula scared the hell out of me. Can you do an example using 20lbs as the weight and 10ft. as the distance dropped. What I need to know is the velosity at impact using those two variables. And I may change variables to get different results.

Can you explain this to me in idiot terms?
 
As long as you work with fairly heavy weights, the role of air resistance will be negligible and so the weight will not matter, only the height will determine the impact velocity.
If you use my above formula, you will find (in feet):
velocity=\sqrt{64.34*height}.
You can use google now. To find the impact velocity at 10 feet just type "sqrt(64.34*10)" (without quotes) into google, hit enter. You will get 25.36. This means your weight will hit the product at a speed of 25.36 feet per second.
Hope that helps.
 
Thanks gamesguru, this helps greatly.
 
Hi there, im studying nanoscience at the university in Basel. Today I looked at the topic of intertial and non-inertial reference frames and the existence of fictitious forces. I understand that you call forces real in physics if they appear in interplay. Meaning that a force is real when there is the "actio" partner to the "reactio" partner. If this condition is not satisfied the force is not real. I also understand that if you specifically look at non-inertial reference frames you can...
I have recently been really interested in the derivation of Hamiltons Principle. On my research I found that with the term ##m \cdot \frac{d}{dt} (\frac{dr}{dt} \cdot \delta r) = 0## (1) one may derivate ##\delta \int (T - V) dt = 0## (2). The derivation itself I understood quiet good, but what I don't understand is where the equation (1) came from, because in my research it was just given and not derived from anywhere. Does anybody know where (1) comes from or why from it the...
Back
Top