Force needed to drive a Keiser Force Machine 10" with 1 hit

[Steve1951]

I am a 66 y/o firefighter that competes in the Firefighter Combat Challenge. One of the evolutions requires the FF to use a Keiser Force Machine. This is a 160# beam on Teflon runners, 2 each 2" x 36", in a stainless still tray. Using a 9# shot hammer we have to drive the beam 5'. The problem has been having the Keisers, 1 in each lane to slide the same every time. I am looking to calculate the force necessary to drive this beam 10" with 1 hit from the shot hammer. The goal is to turn this into a computerized machine, where we can get consistency, but also take into the fact that hitting it again, before it stops moving, you can move it farther with the same amount or force. It has been 40-45 years since I had any physics, but I remember there being formulas to calculate this information. Right now I am looking to see what size the force measuring device needs to be to start. Thanks.

 

[SirLollington]

I might be able to help you here, but I'm not entirely sure what the machine you're trying to drive does, or how it works.
From my understanding, it's just a weight on a rail, and you whack it with a hammer and make the weight slide a certain distance along that rail (Correct me if I'm wrong!)

While this seems pretty straight-forward, there's a few problems here:

First, I'm not sure how to best model the impact on the weight. Is it an elastic collision, or more inelastic (i.e. how does the hammer behave after you hit the weight?)
Second, because the weight is on a rail, the impact first has to break the friction between the rail and the weight. We'd need the coefficient of friction between the rail and the weight to calculate the amount of energy taken to do so.

After that, we should be able to figure out how quickly the weight moves immediately after the impact, and then it's just a matter of knowing how heavy the weight is, and the coefficient of friction between the rail and the weight while it's sliding. This part is pretty straight-forward.

I might be going in a completely wrong direction here, sorry if that's the case.

 

[Steve1951]

This beam we are hitting weighs 160# and sits on 2 Teflon runners that 2" wide and 36" long. This beam sits in a u-shaped tray made of Stainless Steel (S/S). The end of the beam has a flat surface for striking. We use a shot hammer to hit the beam to move it a total of 60" (5'). It simulates chopping on a roof. At an event over the weekend, one of the top competitors was giving a demo on his technique. I noticed that he was able to move it about 10" with just 1 hit. This was from a dead stop position. I tried to give as much info as I could. This included the surface area of contact between the Teflon and the S/S, 144 sq. in.

When it is hit squarely there is very little rebound. When hit at slight angle, thus less contact area, it has considerable rebound. The thing that I observed, was that the angled hit moved the beam the farthest, and there was a considerable difference between the 2 techniques.

I hope this provide more info. If I was asking this 30-35 years ago I might have the ref. book and formulas to make the calculations. If you go to this YouTube video, , at about 1:40 point you will see how the Keiser Force Machine operates.

 

[SirLollington]

For some reason I interpreted the # as "number" and got confused about what it meant. Sorry about that.

Firstly, it's more convenient for me to do this in metric units, I hope that's okay with you. I'll convert the units back at the end
160# = 72.6 kg
9# = 4 kg
10" = 25.4 cm

According to http://www.roymech.co.uk/Useful_Tables/Tribology/co_of_frict.htm, the coefficient of friction between stainless steel and teflon is around 0.04. Unfortunately, there's no data for sliding friction in dry conditions so we'll just have to assume it's 0.04 under all conditions.
The contact surface area doesn't matter, only the force on the contact surface and the coefficient of friction.

The braking force applied while it's moving is 72.6 kg * 9.8m/s^2 * 0.04 = 28.5 N, which causes a deceleration of 0.392 m/s^2. The braking distance is given by d = v^2/2a where d is the distance, v is the velocity and a is the deceleration. Re-arranging this formula for velocity gives us v = √(d * 2a). If we plug our numbers for distance and deceleration, we get an initial velocity of 0.45 m/s.

Unfortunately, I can't find anything telling me how to calculate the energy required to overcome the static friction, so I'll skip that part and hope that it doesn't change the result by much.

Because you said that an angled hit with a lot of rebound moves the weight farther, I think it's best to model the impact as an elastic collision: https://en.wikipedia.org/wiki/Elastic_collision

The formula describing the velocities of both objects before and after the collision is as follows:

m1, u1 and v1 are the mass, velocity before impact and velocity after impact of object 1. m2, u2 and v2 are the mass, velocity before impact and velocity after impact of object 2.

Let's say that the hammer is object 1, and the weight is object 2. We know that u2 = 0 and that v2 = 0.45 m/s, as well as the values of m1 and m2. We don't know u1 and v1, which is a bit tricky, because we only have one equation but two unknowns. However, we also know that the collision was elastic, so we have conservation of kinetic energy. This can give us our 2nd equation: 0.5 (m1 * u1^2 + m2 * u2^2) = 0.5 (m1 * v1^2 + m2 * v2^2)
Below is my working for solving this system:

For conservation of momentum:
4 kg * u1 = 4 kg * v1 + 72.6 kg * 0.45 m/s

Conservation of energy:
0.5 (m1 * u1^2 + m2 * u2^2) = 0.5 (m1 * v1^2 + m2 * v2^2)
m1 * u1^2 + m2 * u2^2 = m1 * v1^2 + m2 * v2^2
4 kg * u1^2 = 4 kg + v1^2 + 14.46

Unfortunately, the 2nd equation has square terms. I'll solve them using the substitution method - to do this, first I'll re-arrange the conservation of momentum formula to give me v1:
(4 kg * u1 - 72.6 kg * 0.45 m/s) / 4 kg = v1

Now I'll substitute that for v1 in the formula for conservation of energy:
4 kg * u1^2 = 4 kg + ((4 kg * u1 - 72.6 kg * 0.45 m/s) / 4 kg)^2 + 14.46

The result is a formula with one unknown. I struggled with it for a bit, but ultimately decided a computer might be better at solving it than I am:
https://www.wolframalpha.com/input/?i=solve+4+*+u1^2+=+4+++((4+*+u1+-+72.6+*+0.45)+/+4)^2+++14.46+for+u1

We see 2 solutions there. Of course, because u1 is technically a vector, it must have a positive length - so we choose the solution that is a positive number, and we get 3.26 meters per second. This is the speed the hammer moves before the impact.

That's about 10.7 feet per second. In the video it looks like the hammer is accelerated to that speed in about half a second (may have been different in that demo, I don't know that), which would require a force of 26.08 Newtons or, roughly 5.86 lbs to be applied throughout the entire acceleration.

Hopefully you found this useful

 

[jack action]

4 kg * u1^2 = 4 kg + v1^2 + 14.46

Just so you know, there is an error in that equation that was carried on throughout the calculations. It should read:

4 kg * u1^2 = 4 kg * v1^2 + 14.46

Anyway, from the wikipedia page about elastic collision, there are 2 equations to find the velocities (same answer as https://www.wolframalpha.com/input/...((4+*+u1+-+72.6+*+0.45)+/+4)^2+++14.46+for+u1):

Setting $u_2 = 0$ and $v_2 = 0.45\ \text{m/s}$, you will get $u_1 = 4.31\ \text{m/s}$ and $v_1 = -3.86\ \text{m/s}$. And the force will be 34.47 N or 7.75 lb.

 

[SirLollington]

Just so you know, there is an error in that equation that was carried on throughout the calculations. It should read:

4 kg * u1^2 = 4 kg * v1^2 + 14.46

Whoops. I missed that. Thanks for correcting my mistake for me.

 

[JBA]

When it is hit squarely there is very little rebound. When hit at slight angle, thus less contact area, it has considerable rebound. The thing that I observed, was that the angled hit moved the beam the farthest, and there was a considerable difference between the 2 techniques.

With regard to the above, a shot hammer is designed to evenly distribute the impact across the hammer face to in one case prevent possible fracturing and shrapnel when impacting two hard surfaces. While I do not know the specific design of the hammer used in your case, based upon your observation about the rebound of the off angle strike it would appear that there is a rigid structure surrounding the primary hammer striking face. The result of the harder surface impact, while it does not necessarily increase the total delivered energy of the blow, is that the peak force of the strike is higher. As for the observed positive effect, assuming the bar is at a full stop prior to the impact, then this spike force might have a positive effect on overcoming the initial higher the static friction at the initial launch of each bar movement.

 

[Dr.D]

Firstly, it's more convenient for me to do this in metric units,

The system of units should not make any difference at all, as long as a consistent system is used. To convert unit unnecessarily is a bad practice because it introduces additional opportunities for error. In this case, the problem was stated in US Customary units, and could easily be solved in that system which is entirely consistent.

 

[CWatters]

Would it be possible to require each FF to have one go on each lane? That way if they are slightly different it evens out in the end.