(I was informed on the Physics board that this is an Engineering question, therefore they closed it there, and I moved it here)

First, I have next to zero formal physics training... I opted for all math & electronics electives back in my 1970s high school curriculum.

Goal: My goal is to create a replacement for a "vertical drop test" by accelerating an impact mass in the horizontal plane.

The existing Vertical Drop Test machine that I want to replace drops a small product 1 meter onto a steel plate, but the orientation of the product is not predictable, and it needs to be.

By creating a Horizontal impact machine to substitute for the vertical drop, the product can then be pre-positioned on the machine so that the impact happens at a precise location on the product being tested.

My main challenge is understanding how to calculate what mass to accelerate at the "product", and at what velocity.

My example follows: Assuming the object being dropped vertically is a 100g solid plastic ball. I calculate the impact speed of a 1 meter vertical drop to be 4.429 m/s, resulting in 0.98 J kinetic energy.

On my proposed horizontal impact machine, I would have the same ball resting on a horizontal shelf (without constraint), and impact it horizontally with a given mass/velocity. The "ball" will then move away from the point of impact, just like it would bounce when dropped on the floor. If that horizontal impact mass is 250g (with essentially no deflection) moving at 2.8 m/s (which is still 0.98 J Kinetic Energy, I believe), are the impact forces equivalent to the vertical drop regardless of the "give" (deflection) the "plastic ball" may have?

I know that if the "ball" were stationary, and I moved an infinite mass (the floor) up towards the ball" at the same speed as the ball being dropped vertically, the impact would be the same. (right?)

If I use a lower impact mass, I would need a higher velocity to equate to the same impact Force, is that correct?

Does the impact mass need to be many times (10x or greater) the mass of the product being impacted, in order to cause the equivalent impact Force?

I love to learn, but again, my physics background is minimal... I'd appreciate any lay-language help, with formulas (or even a link to a web site with a calculator for such a problem).


A bit more "detail". This is for testing watch buttons... it's called a "Button Focused Drop Test", and is not an ISO regulated test.

The problem with the "normal" way of performing this test, is that the machine releases the "sample" as it's falling, but well before it impacts the target... which in this case is a steel plate. The watch almost always tends to "tumble", making the impact point/position unpredictable. In addition, when watches are "square", and have multiple buttons on a flat side are tested, it's very hard to isolate a specific button. Also... once it's dropped and impacts the steel plate, it bounces elsewhere, possibly incurring additional damage not related to the initial impact.

If the watch is laid on a horizontal "shelf", and the impact device is pre-aligned so that the impact position is precisely controlled, and once the impact occurs it falls into a padded receptacle, the initial impact is all that happens... and the results are extremely predictable.

  • 1
    $\begingroup$ Depends on what you need. Seems to me thatas a simulation its not entirely the same if the impaxt speed is not the same.But that might be tricky because we dont know the impulse. Why would ot be imposible to engineer the drop test to allways drop on the desired orientation? $\endgroup$
    – joojaa
    Jul 20 '20 at 21:28
  • $\begingroup$ Are you looking for those swinging demolition balls? $\endgroup$
    – Solar Mike
    Jul 21 '20 at 6:19
  • $\begingroup$ N.B. Tests done with this rig will not be accepted where an eg ISO drop test is required for regulatory purposes, even if it is broadly equivalent in the damage produced. $\endgroup$ Jul 22 '20 at 7:57
  • $\begingroup$ Don't forget that friction will impact the test in horizontal positioning whereas an object in free fall only has air displacement for almost zero resistance. $\endgroup$
    – Rhodie
    Jul 26 '20 at 23:20

Freefall friction-less velocity of an object falling 1 meter is

$$ V= \sqrt{2hg}= \sqrt{2g}=4.427m/s $$

This velocity decreases to zero in a short time after the crash, and unless your ball is not a paste it will bounce back up with the same speed in a ping pong manner falling down again. Let's assume you use a horizontal weight, M, n times heavier than your plastic ball.

As per Newton's law of conservation of moments in a collision: $$\Sigma mv=0 \rightarrow m*0+M*V_{initial}=M*V_{final}+mv_{final}$$ And we have sum of kinetic energies before and after contact equal.

If your ball would not bounce on the vertical drop then the weight collides with and becomes one with the ball:

$$M*V_{initial}= (m+M)*4.427m/s$$

M has to be moving initially faster to absorb the inertia of the ball and together they move at 4.427m/s


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