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1/25/2018 7:38:29 AM
Posted: 4/8/2002 8:02:13 AM EST
I have gotten 3 responce from 3 different engineers. Each one came within a couple HP of each other. It seems this all boils down to the "time " part of the equation. A very large amount of power over a very very short time.. Im sure the results from a .223 would be just as amazing. These results are way higher than I ever thought. Given that you want to know horsepower (hp) these are like the proverbial apples and oranges. Power is the rate of change in the energy something posseses. Or mathematically (change in energy)/(change in time) or dE/dt where calculus is concerned. It can be estimated for the acceleration of a bullet given two not so great assumptions. 1. That the rate of acceleration of the bullet traveling down the bore is constant. Or in other words, lets take your example. If for a 31" barrell, the muzzle velocity is 2800 ft/s, then when the bullet is 15.5" (half way) down the bore it's velocity is 1400 ft/s, etc..... for estimation purposes this is not that bad of an assumption. and 2. If we disregard the energy lost in the process of propelling the projectile down the bore due to friction, (which means that we will underestimate the power by at least 20% (Although I am totally guessing on that one). So I believe you are asking "What is the power generated in propelling the projectile down the bore?" It can be roughly estimated by dividing the change in energy by the amout of time the energy change occured. So here goes, the change in energy is 12000 ft-lbs (since it starts with zero kinetic energy) and if we use the constant rate of acceleration assumption I get 0.00185 seconds for the time required. So 12000(ft-lbs)/0.00185(s) = 6,500,000 (ft-lbs)/s Which are odd units of power. If we use the metric unit for energy (the joule 12000 ft-lbs equals 16,270 J) this value of energy divided by the same amount of time gives us 8,795,000 Joules/s Which is the units of Watts. and 745.7 watts equals one horsepower (hp) so dividing our answer in watts by 745.7 gives us the astonishing value of about 11,825 Horsepower!!!!!!! And yes this is a very high power output, and if you consider the power output of the powder, It is underestimating the value due to the energy lost due to friction (as well as heat energy lost to the "cooler" sides of the bore from the hot propellant gasses. But remember that this power is output only for a small amout of time. You can think of it this way, The total amout of ENERY released from approx. 225 grains of rifle powder is probably about the same as that released from burning about a pint of gasoline in your lawnmower mowing your yard. However since that enregy is released over a 30 to 60 minuite time frame the power output is MUCH less, say 4 to 5 hp (typical mower engine power). Wow, I guess I turned this into an entire physics lecture or something, hope this helps Bill
Link Posted: 4/8/2002 8:34:35 AM EST
I'm no math expert, but I thought if you knew the weight, the distance traveled, and the time it takes to move the weight that distance, you could calculate horsepower... Very plain simple math... (Of course I don't remember how) Something like: 100lbs moved 10ft vertical in one minute= 1 horsepower... I know that's wrong, but that's what princile I was taught many years ago... If you know the bullet weight is 750gr, and it travels 2700ft in 1 second, you should be able to somehow convertthe energy to horsepower... 'Sorta like drag racing... If you move 3100lbs a distance of 1320ft in 8-1/2 seconds, thats roughly 1000 Horsepower... There are power/speed calculators and computer programs that will give EXACT horsepower with these known inputs... Just tring to simplify things, and offer another angle to approach a HP number... From what I understand, it's fairly simple math, but not simple enough for my more simple mind [:D]
Link Posted: 4/8/2002 10:42:01 AM EST
I think 1hp is the ability to move 550lb 1 foot in 1 minute, or some combination thereof. I just have a hard time believing that a .50 bullet is propelled by a force powerful enough to move 550lb 10,909 feet in a minute! I also think that part of the problem is HP is a unit of measure that is dependent upon rate (1hp=550lb 1ft/min or 550lb 0.2in/sec, 0.5hp=550lb 1ft/2min or 275lb 1ft/min) but lbf ft is basically a "snapshot" of the energy being applied to an object at a certain time. when I torque a cylinder head bolt, I'm not applying 100lbf ft for a length of time, I'm just increasing the force I apply until it reaches a certain value. Once I'm at 100lbf ft, I can stop right away and go to the next bolt, or hold 100lbf ft for 3 days - it won't change how much force I'm applying. Now, I will be applying many more hp if I apply 100lbf ft against that head bolt for 3 days (assuming the bolt was also trying to apply 100lbf ft on me at the same time), than I would be if I applied the same 100lbf ft against the head bolt for, say, 0.1 sec. I think the 3 engineers are most likely measuring cumulative hp over time, but once the bullet leaves the barrel there's no longer a thrust force acting on it. Again, I could be VERY wrong since physics is far from my specialty, but hp is generally measured as peak hp, and not the total hp required to move a car down the track or a bullet out the barrel and downrange.
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