Turbulent Turtle
F-TR competitor
In an earlier thread, I was involved in a short debate about how quickly the spin rate of a bullet at the muzzle decays during the flight and what, if any, effect this may have on increased or decreased bullet stability.
We were veering off the thread’s subject and I didn’t have an opportunity to continue the discussion due to business travel and all that. Now that I have a bit of time, I’m starting a thread specifically on spin decay so as to not hijack someone else’s thread.
This was also prompted by an article in the current issue of Military History that came to my smartphone just before boarding the return flight last week. The article, in the May 2015 issue, is entitled “Paris Under The Gun†and it talks about the German guns that shelled Paris from about 80 miles away starting in March, 1918.
The article describes the guns, how they were made and the firing sequence. It is truly amazing and about as complicated as it can get for a piece of artillery. The article also explains that the time of flight was something in the order of 177 seconds from 80 miles away and that the guns were elevated at greater than 50 degrees to send the shells into the stratosphere to reduce drag and have them travel further.
On a personal note, the article has a sidebar about Dr. Gerald Bull and his HARP gun and how he reconstructed the mathematical model behind the Paris guns. I never met Dr. Bull but I did go to school with his sons in the 1960s. He was assassinated in Belgium, ostensibly by the Mossad because he was working on a supergun for Saddam Hussein after being ejected from Canada.
At any rate, I was thinking that if we do see spin rate decay in the few seconds of flight from a rifle, we would probably see a lot of decay in three minutes and the shell might become erratic upon reentry into the lower atmosphere.
So, I wanted to see if it was possible to model spin rate decay, regardless of the effect on the flight of the projectile. I believe we all saw the Youtube videos showing a handgun bullet spinning madly and for quite a while when fired in snow and ice. Interesting and it shows the length of time the spin can continue even with friction from the ice to slow it down. But what about pure flight.
Let’s take something simple; a .308 bullet in a 1:12 twist with an MV of 3000 FPS. The bullet will spin at 180,000RPM or 3,000 RPS (rotations/second). I’ll use that last number because it’s easier. The angular velocity of the bullet is spin time radius and in this case, this works out to 462 inches per second for a .308 bullet with a radius of .154 inch at 3000 RPS. Calculating the kinetic energy of the spinning bullet becomes more complex because of the shape of the ogive and to a lesser extent, the shape of the boat tail. The regular equation that works well for cylinders is overstating the kinetic energy contributed by said ogive and, to a lesser extent, the boat tail. I would need to find formulae for the KE of a cone (ogive) and approximate the KE for the boat tail. I’m not even sure if it’s even important in this discussion and I’m afraid of creating a new unit of measure, the inch-grain.
Now, once the bullet leaves the barrel, the only thing that will retard the spin is whatever resistance (force) is presented by the air around it catching into something on the bullet. It was postulated that the rifling marks engraved by the lands of the barrel would be the sail or rather, the air brakes slowing down the spin. So that is what we need to measure and then see what effect it could have on the spin. In a bullet the bearing surface dictates the length of the rifling mark; the ogive and the boat tail would not have any of those marks. So, let’s stipulate that our .308 bullet (a JLK .308 155 gr to get to 3000FPS) measures 1.250 inch of which the bearing surface makes up .320 inch. Grooves in a barrel are usually .004 inch deep so in a 4 groove barrel we would have 4 airbrakes about ¼ the length of the bullet and .004 inch tall. Each groove represents .00128 square inches for a total of .00512 square inches of airbrake.
Ok, so we have the braking surface and the angular velocity of the bullet. What I do not know is how to relate the two in a supersonic flight regime, especially given the cone shape of the ogive. Since the bullet ogive creates a shock wave at supersonic speed, I’m not sure what the air pressure is for the .320 inch of the bearing surface, but I think it’s less than ambient density or else airplanes would not fly. Also, I keep thinking the engraving has zero to negligible effect because it might just be shallower than the boundary layer of the air even without the bullet going supersonic.
Well, that’s as far as I have taken it for the moment, I need an aspirin.
We were veering off the thread’s subject and I didn’t have an opportunity to continue the discussion due to business travel and all that. Now that I have a bit of time, I’m starting a thread specifically on spin decay so as to not hijack someone else’s thread.
This was also prompted by an article in the current issue of Military History that came to my smartphone just before boarding the return flight last week. The article, in the May 2015 issue, is entitled “Paris Under The Gun†and it talks about the German guns that shelled Paris from about 80 miles away starting in March, 1918.
The article describes the guns, how they were made and the firing sequence. It is truly amazing and about as complicated as it can get for a piece of artillery. The article also explains that the time of flight was something in the order of 177 seconds from 80 miles away and that the guns were elevated at greater than 50 degrees to send the shells into the stratosphere to reduce drag and have them travel further.
On a personal note, the article has a sidebar about Dr. Gerald Bull and his HARP gun and how he reconstructed the mathematical model behind the Paris guns. I never met Dr. Bull but I did go to school with his sons in the 1960s. He was assassinated in Belgium, ostensibly by the Mossad because he was working on a supergun for Saddam Hussein after being ejected from Canada.
At any rate, I was thinking that if we do see spin rate decay in the few seconds of flight from a rifle, we would probably see a lot of decay in three minutes and the shell might become erratic upon reentry into the lower atmosphere.
So, I wanted to see if it was possible to model spin rate decay, regardless of the effect on the flight of the projectile. I believe we all saw the Youtube videos showing a handgun bullet spinning madly and for quite a while when fired in snow and ice. Interesting and it shows the length of time the spin can continue even with friction from the ice to slow it down. But what about pure flight.
Let’s take something simple; a .308 bullet in a 1:12 twist with an MV of 3000 FPS. The bullet will spin at 180,000RPM or 3,000 RPS (rotations/second). I’ll use that last number because it’s easier. The angular velocity of the bullet is spin time radius and in this case, this works out to 462 inches per second for a .308 bullet with a radius of .154 inch at 3000 RPS. Calculating the kinetic energy of the spinning bullet becomes more complex because of the shape of the ogive and to a lesser extent, the shape of the boat tail. The regular equation that works well for cylinders is overstating the kinetic energy contributed by said ogive and, to a lesser extent, the boat tail. I would need to find formulae for the KE of a cone (ogive) and approximate the KE for the boat tail. I’m not even sure if it’s even important in this discussion and I’m afraid of creating a new unit of measure, the inch-grain.
Now, once the bullet leaves the barrel, the only thing that will retard the spin is whatever resistance (force) is presented by the air around it catching into something on the bullet. It was postulated that the rifling marks engraved by the lands of the barrel would be the sail or rather, the air brakes slowing down the spin. So that is what we need to measure and then see what effect it could have on the spin. In a bullet the bearing surface dictates the length of the rifling mark; the ogive and the boat tail would not have any of those marks. So, let’s stipulate that our .308 bullet (a JLK .308 155 gr to get to 3000FPS) measures 1.250 inch of which the bearing surface makes up .320 inch. Grooves in a barrel are usually .004 inch deep so in a 4 groove barrel we would have 4 airbrakes about ¼ the length of the bullet and .004 inch tall. Each groove represents .00128 square inches for a total of .00512 square inches of airbrake.
Ok, so we have the braking surface and the angular velocity of the bullet. What I do not know is how to relate the two in a supersonic flight regime, especially given the cone shape of the ogive. Since the bullet ogive creates a shock wave at supersonic speed, I’m not sure what the air pressure is for the .320 inch of the bearing surface, but I think it’s less than ambient density or else airplanes would not fly. Also, I keep thinking the engraving has zero to negligible effect because it might just be shallower than the boundary layer of the air even without the bullet going supersonic.
Well, that’s as far as I have taken it for the moment, I need an aspirin.
