**Updated: Please see additional information at the bottom.**

Hey Folks.

So, in the real world, I’m a mechanical engineer, and this same discussion has come up a number of times in a number of places and I took the time to go through some of the math. This is going to be a largely technical post, so it’s not the most exciting read. But, here’s my answer:

Are roller guns any advantage, and if so, why?

I see this question argued a lot, and people commonly talk about impulse and counteracting forces and all kinds of wild and woolly things. They’re definitely barking up the right tree here, but they’re making it a bit more complicated than it needs to be.

The most important bit where people are getting confused is with the tricky concept of impulse. The technical definition of impulse is the integral of force with respect to time, which means if force is constant, it’s just force times time. However, in this situation, the time it takes the shaft to leave the gun will not be the same if the final velocity is the same. Now, WORK is actually a bit more useful in the comparison, because then we don’t have to deal with time, because work is the integral of Force with respect to distance, or if force is constant, force times distance. This greatly simplifies our analysis, because we can look directly at kinetic energy.

Ok, now that we have that set up, we can look at two different situations very simply. We have gun A and gun B. Both guns have exactly the same mass, and are exactly the same length. Both fire the EXACT SAME SPEAR (with no line, because who wants to switch lines and all that nonsense). When they fire that spear, their final velocity is identical. It is identical, because we’ve spent years tuning the two guns to shoot at identical velocities, right. Because we have that kind of time. Maybe we live in the folk’s basement and just yell about meatloaf whenever we’re hungry, but that’s not important right now.

Now, Gun A is a conventional 2 band gun with bands stretched exactly enough to produce velocity X, and lets say it pulls for 2/3 of the barrel. Gun B is an inverted roller with the bands completely on the back side of the gun, for simplicity’s sake, and it pulls for the entire length of the tube. We can all agree that this setup is possible, yes? These guns fire a spear identically, but the way they do it is different.

Now here’s a really important bit: What is recoil? I’ll define recoil as the acceleration of the gun backward into your hand that must be countered by a force from your hand. If you want to argue that definition, I’ll listen, but I think it’s pretty fair. Now, thanks to F = MA, we know that that acceleration is equal to the sum of the forces on the gun divided by the mass of the gun (A = F/M). Since both guns have the same mass, we can now be sure that recoil is affected ONLY by the sum of the forces on the guns. So, since the force of the bands contracting isn’t constant over the pull, then the most important part we can assume is the peak force. Do we agree on this? So, if we prove that the peak force on Gun A is greater than Gun B, then we also prove that the recoil on Gun A is greater than on Gun B. Agreed?

So, what are the forces on the guns? They are the force of the acceleration of the shaft, the force of the acceleration of the bands, and the force coming through your hand as you try to stop the recoil. On Gun A, the bands accelerate in the same direction as the shaft, so the sign is positive. On Gun B, the bands accelerate in the opposite direction, so it’s negative.

Now, this means the bands do account for some percentage of the total force in each situation, but how much? This is actually where my analysis proved my initial gut instinct wrong. Well it turns out, proportional to the mass ratio between the bands and the shaft as well the percent elongation. This is a first order response in both situations, meaning there are no exponents. If I double one parameter, the outcome doubles. If the bands are half the mass of the shaft and the center of mass of the bands moves exactly half the distance the center of mass of the shaft in the same time, then the MAXIMUM force these bands can produce is 1/4 the force the shaft produces. This is higher than I initially believed it to be, which caused me to actually go a bit more into detail than I wanted to, and I actually looked into the integrals, leaving rigid body kinesthetics behind and drifting into the massive wasteland of math beyond it. What i found is that simple rigid body kinesthetics actually DO apply for this situation, even though it isn’t a rigid body. Since the force increases linearly with stretch below 500% elongation, we actually CAN just track the center of mass and use the average force and end up with the the proper work. This is ONLY true if the bands are ENTIRELY on one side of the gun or the other. Otherwise, it gets much more complicated. So, traditional rollers where the band makes a turn changes some things and some integrals with respect to time or distance get uglier, but it can still be instantaneously approximated fairly simply. I also weighed a couple shafts and bands, and found that a mass ratio of .5 is pretty reasonable. So, this is a swing of +/- 25% of the kick from the spear, assuming the peak force of the bands is the same.

So, part one rules that the force of the recoil of the conventional gun will be 125% the kick of the spear leaving the gun, and the roller gun will kick with 75% of the force of the spear, proven by only a sum of the forces. That’s a far more significant cut than I initially believed, but we’re not done.

Now we have to compare the acceleration of the spear, since they’re not accelerating identically. Since we know the work is identical, and we can model work as Force x Distance, and we know that Distance A is .6667 of Distance B, since the conventional gun, Gun A, only pulls for 2/3 of the length of the barrel, while Gun B pulls for the full length. This leaves us with the following

>Wa = Wb

>Fa*Da = Fb*Db

>Da = .6667*Db

>Fa*(.6667*Db) = Fb*Db; Solve this for Fb

>.6667*Fa = Fb; Distances cancel, leaving only the forces and the ratio between distances.

Thus, the average force in Gun B only has to be 2/3 the average force in gun A. Since in both cases, these forces increase linearly with respect to distance, we also know that this ratio is true of peak force at the initial position. This is also the force required to load it.

So, we now have two ratios. Both are first order responses, so we can actually combine them. Let’s give the force of the acceleration of the spear from Gun A an arbitrary value of 100 N. This gives Gun A a total kick of 125 newtons, since we know that the bands add 25% of the force of the spear in this situation.

Now, for Gun B, the force of the spear is only 66.7, because it’s 2/3 of the force in Gun A as we proved a second ago. Then, we have to subtract the force of the bands, because they’re going in the opposite direction. Now, we only get to take out 25% of 66.7, not 100, because that’s the force on THIS spear. So, we end up with a final kick of 50.025 N for the EXACT SAME SPEAR moving with the EXACT SAME VELOCITY. Using this arrangement cuts the kick by almost exactly 60%. This is all physics here, no guesswork.

So, what else does this tell us? Well, it tells us that the higher our mass ratio is, the more our bands are adding to our kick. It tells use that kick increases with mass of the spear or mass of the bands. It tells us that if the bands make any corners, such as a standard roller gun, this will take away from the counteracting force of the bands if they’re all on the back. So, a conventional roller with about the same amount of band on the top and the bottom when the bands are stretched will have more kick than if the bands are all on the bottom. However, since the top part will be accelerating much faster than the bottom, you can treat this as two separate bands initially, with the one on the back moving half the speed of the one on the front. This will end up with the bands adding a kick of 12.5% the kick of the shaft alone. You can just trust me on that, I’m not going through it any more detailed.

Now, just in case we don’t all agree that the work done by both guns is equal, let’s look at some math, just so no one comes back with “nuh uh”.

Terms

>M = Mass (units kg)

>V = Velocity (units m/s)

>D = Distance (units m)

>KE = Kinetic Energy (units kg*m^2/S^2 or Joules)

>KE = 1/2MV^2

>W = Change in KE

>Delta KE = 1/2M2*V2^2 – 1/2M1*V1^2

>M1 = M2; The spear doesn’t lose any mass.

>V1 = 0; It wasn’t moving at first

>Delta KE = 1/2M*V2^2; Second term goes to zero, since V1 is zero.

Therefore, we know that work done on the spear is a function of ONLY final velocity. So, it doesn’t matter how the spear go there, the net work done on it is exactly the same if the final velocity is the same. We already proved that the force required to make this work is directly proportional to the distance over which it acts.

Now, this is getting really long, but here’s a bit about this gun in particular.

Pulleys are force multipliers. So, the tension in the line going around the pulley is EXACTLY half the tension in the band pulling on the pulley itself. All our previous work just showed that the force required for a roller is 2/3 the force of a conventional gun, but now I’m saying that the pulley halves the force coming from the band, so, what’s the advantage? Now the force we need in the bands is twice what it would be, or 4/3 of the conventional. This should be bad. However, the other thing the pulley does is cut the travel of the band in half when compared to the line, so the acceleration of the COM of the bands goes down to a quarter of that of the spear. Additionally, our mass ratio is probably going to have to go up, but let’s pretend it doesn’t. So, acceleration cut in half, mass ratio stays the same at .5, so now the bands are only producing 1/8th the force of the shaft, but since they’re going the other way, this is worse than the roller we looked at earlier. If we set this gun up to match the initial conditions of Gun A and B, this one is actually going to kick 58.36 N. So what’s the advantage? Well, we can pretension this one and all the sudden the slop of our force over the distance of the pull is half the slope that it would be on Gun B, even if we did an identical percent pretension. Since our travel is so much shorter, we can stay in that high stretch range longer with much shorter bands, which keeps our mass ratio between band and shaft down, while making our force closer to constant. Without this pretension, there is actually a disadvantage to having the pulley. So, essentially, whether the pulleys are good or bad all comes down to setup and tuning.

Anyway, this is really long and I can imagine it being a bit of a boring read, but the info is there. If there’s any part of it that you’d like to argue with, please let me know. Thanks for reading.

**Update**

So, I’d like to clarify a few things and then sum it all up here, since i’ve been getting some argument here and there and a few questions.

First, I’m not using F = kx because it doesn’t actually matter, as the math points out. First, k is a spring constant, not a material property. It’s a function of Young’s modulus and geometry. A longer spring has a lower spring constant, but force for a given band diameter IS a function of % elongation. So, basically, you can get any peak force you want out of any gun, within the limits of the stresses the trigger mech and barrel can withstand. All you have to do is cut your bands shorter or add more bands. So, comparing just the peak force of two different types of guns tells you nothing, because they are free variables.

Second, stored potential energy isn’t even that important, because similarly, you can load up any gun with any amount of potential energy you want. All you have to do is cut your bands or add more.

Third, what really matters is the ratio between peak force and total work. THAT is where the rollers all the sudden have an advantage. They can produce greater work at a lower peak force. This is again true when you go from rollers to pulleys with pretension.

The ideal gun produces force as a step function. It begins producing force the instant you pull the trigger and maintains constant force until the spear leaves the gun, imparting the desired total work using the least peak force. If you were to graph Force over Distance of this gun, you would get a perfect rectangle on the graph. For a conventional gun, you get a triangle. With a conventional roller of the same type, you get a triangle with a base that is ~125-150% that of the conventional gun. With a pretensioned pulley gun, you get a trapezoid that looks like a much longer triangle with the end cut off. If all of these have identical areas (work), then the trapezoid will have the lowest peak force, and therefore the lowest recoil resulting from the spear leaving the gun. This is why it appears that you can get more power for the length of your gun out of a roller, because it produces more work for a lower peak force.

Finally, I’m not saying that rollers are categorically better and we should all go out and buy one and burn all our conventional guns. The point of this post is to clear up the physics involved and show the reasons people are investigating the concept. It also provides a basis to tune a roller gun from, since we can look at the physics and try to find out where we’re going to get the most gain over a conventional gun. There are plenty of disadvantages to a roller, such as size, bulkiness, complexity, additional lines running all over hell, and they’re a bit awkward to load, especially if you’re short.

FINAL VERDICT: Roller guns provide greater work for less peak force than a conventional gun, allowing reduced recoil for a given shaft kinetic energy. This is the findings of the physics involved. Whether or not this is worth it depends greatly on the setup, and there will be a point of diminishing returns where reducing recoil adds little to no value.