Rifle: A machine rated in horsepower

I'm afraid you will have to copy and paste it here. The only MAJOR error I've seen is that you still seem to think that you are "doing work" to the bullet 2997.8 ft after it has left the barrel...you are not. You "Did Work" to it ONLY for the 26" (or so) that it was IN THE BORE. Afterwards it is not being "worked upon" by you.
 
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Vibe said:
Using the 4195 ft-lb muzzle energy quoted before, and the 0.001711 sec. barrel time, results in 2451782.583 ft-lb/sec

divided by 550 ft-lb/sec - I get 4457.786515 HP

Working backwards
.668 Hp = .668 * 550 ft-lb/sec = 367.4 ft-lbs/sec

367.4 ft-lbs/sec * 0.001711 sec =0.6286214 ft-lbs muzzle energy = pretty anemic.
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From what is at this moment, post # 27 from you.

Now.
The only MAJOR error I've seen is that you still seem to think that you are "doing work" to the bullet 2997 ft after it has left the barrel...you are not. You "Did Work" to it ONLY for the 26" (or so) that it was IN THE BORE. Afterwards it is not being "worked upon" by you.
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You say, this is not a MAJOR ERROR?????????????????????????????????

Please, let's start over here and don't make me copy and paste every instance I told you this is wrong because we don't have enough space on the internet.

Now that we know the muzzle energy has been used grossly wrong, I'll admit that the hypothetical barrel length does alter the numbers but for OUR DISCUSSION, we were using average values to make the math easier.

As far as that causing any appreciable error as you keep grasping at straws with, gravity will only affect the bullet by 9.8 Meters as we are using one second for the bullet to move 3000 feet. So, it will be 296x.x feet of travel instead and I SAID THIS LONG AGO.

So, if our barrel is one foot long, or 3000 feet long, that was not ever defined and does not change the answer does it.

Some time ago, you made a statement about my not being able to understand my own reference. Let's see, would you like to correct that now?

In the above example, you arrived at
2451782.583 ft-lb/sec
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NOw, is that number correct? Is that considered a MAJOR error? Yes or no.

Let me ask a better question. If your bank inadvertantly took a check of yours for $3000 (the amount you wrote the check for) and charged your bank account $2,451,782.58, would you ask them to reconsider?
 
H versus V

I agree that horizontal versus vertical will not make a difference in the final answer. However, I can see that the difference in direction is causing some difficulty.

The other thing that seems to be causing problems is related to the SAE units of measure. In particular, pounds of force as compared to pounds of weight.

Normally, I prefer SAE units of measure also. However, in this case, might it be more transparent to use metric units? We all know that a Kilogram is not a Newton, but it's not always apparent that a pound isn't always a pound.

The units have to come out right. That's the main thing that got me through Thermodynamics and Heat Transfer classes in college. I knew that if the units worked out right, at least there was a possibility that I had the correct answer. If the units didn't worrk out, there was no way that my answer was correct. (Metallurgical Engineering students have some overlapping curriculum with Mechanincal Engineering students).

Anyway, my suggestions are to agree upon a single direction (up or horizontal) so that the effect of gravity is understood clearly, then switch to metric units of measure to avoid issues with pounds of force and pounds of weight.

SteveM.
 
4Mesh said:
You say, this is not a MAJOR ERROR?????????????????????????????????
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Not only is it not a "major" error....it is completely correct in that it represents the CHANGE in energy from chamber to muzzle upon firing.

I demonstrated this in post#62 as well and even provided all of the math to support it.

4Mesh said:
Now that we know the muzzle energy has been used grossly wrong, I'll admit that the hypothetical barrel length does alter the numbers but for OUR DISCUSSION, we were using average values to make the math easier.
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Not only was it NOT used "Grossly wrong" it's a well documented "short cut" to the correct value of energy CHANGE between chamber and muzzle. What happens past the muzzle is of no consequence to the issue. At least not the issue of Horsepower.




Is that MY avatar I'm hearing?
 
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Riddle me this.

Is 4000 horsepower applied for 0.001 seconds the same amount of work as 4 horsepower applied for 1 full second?

SteveM.
 
Ok Vibe,

Using the assumptions in my post #12 that seemed to be good enough for you to take a first stab at this, how about you do a proof on this. You have supported this for a rather long time now. Please re-read the thread if you missed or need to see all the times I've asked you to support why you are using energy to calculate power, really just questioning the end result.
Vibe said:
Using the 4195 ft-lb muzzle energy quoted before, and the 0.001711 sec. barrel time, results in 2451782.583 ft-lb/sec

divided by 550 ft-lb/sec - I get 4457.786515 HP
Click to expand...
In there, you used my hypothetical 210 grain bullet in a 300 win mag and a velocity of 3000fps included in the energy, but also used as an estimate of distance for work (that is within roughly 99% of the actual value if you use only the barrel time and disclude drag as that has nothing to do with what HP the gun made. Another point both of us have made multiple times. No other reference mentions 4195 ft of energy so it was my example you were "correcting" shall we say.

Now. Discluding other insignificant or irrelavant/ignored forces as you did in this example, do you still say the 2451782.583 ft-lb/sec power is an accurate estimate for this hypothetical situation and accurate enough for this discussion? Yes or no.

Now, Is it within 100 THOUSAND ft-lb/sec of the truth, yes or no?
Is it within 500 THOUSAND ft-lb/sec of the truth, yes or no?
Is it within A MILLION ft-lb/sec of the truth, yes or no?
Is it within 2 MILLION ft-lb/sec of the truth, yes or no?
 
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210 grain * 1 pound/7000 grain * 1 slug/32 pounds = 0.0009375 Slug
(3000ft/sec)^2=9000000 ft^2/Sec^2

1/2MV^2= 1/2*0.0009375 *9000000 Slug-ft/sec^2 -ft

1/2MV^2= 4218.75 ft-lbs energy

Energy in chamber = 0
Energy at muzzle = 4218.75

(Energy at muzzle)-(Energy in chamber) = 4218.75 ft-lbs = work done

How much time did it take to do the work?
Again I'm going to assume a 26" barrel
Velocity at muzzle = 3000ft/sec
Velocity in chamber = 0
Assuming constant acceleration V(ave)=(Vmuzzle-Vchamber)/2=1500ft/sec
distance =Vt
so
T=d/V =(26" * 1ft/12")/(1500ft/sec)=2.166666667/1500=0.001444444 seconds

Work done/time to do the work= 4218.75/0.001444444=2920673.077 ft-lbs/sec

Work done in terms of HP
(2920673.077 ft-lbs/sec)/550ft-lbs/sec=5310.314685HP

Now. Discluding other insignificant or irrelavant/ignored forces as you did in this example, do you still say the 2451782.583 ft-lb/sec power is an accurate estimate for this hypothetical situation and accurate enough for this discussion? Yes or no.
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It was a bit low. The 0.001711 sec. barrel time get's it a bit closer though, but I don't really know where that came from.
 
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Vibe said:
(Energy at muzzle)-(Energy in chamber) = 4218.75 ft-lbs = work done
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Wrong again. Here you go with energy = work. energy does not equal work.

Didn't I just get done spending a day teaching you that these two are not the same thing? I think your answer was, "Why would they be?"
 
4Mesh said:
Wrong again. Here you go with energy = work. energy does not equal work.

Didn't I just get done spending a day teaching you that these two are not the same thing? I think your answer was, "Why would they be?"
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Energy does not equal work...but the CHANGE in energy DOES.
Work07.gif

* The Energy transferred into a system by the action of a Force is the Work done on the System.


According to the work-energy theorem if an external force acts upon an object, causing its kinetic energy to change from Ek1 to Ek2, then the mechanical work (W) is given by:
7e9050971f758b530cab178701beff0d.png
 
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Somebody help me find my error.

Help me find any errors with this methodology (using metric units and expressing power in terms of Newtons*Meters/Seconds) :

210 grains = .129 Newtons

26 inch barrel = 0.661 Meters

Time to reach end of bbl = 0.0017 seconds (estimated)

Watts = 0.129*0.661/0.0017 = 50.16 Watts

Since roughly 745 Watts = 1HP,

50.16 Watts = 0.0673 Horsepower


Seriously, I don't see the error in this logic (not saying there isn't any, just that it's not obvious to me).

SteveM.
 
Jetmugg said:
Help me find any errors with this methodology (using metric units and expressing power in terms of Newtons*Meters/Seconds) :

210 grains = .129 Newtons

26 inch barrel = 0.661 Meters

Time to reach end of bbl = 0.0017 seconds (estimated)

Watts = 0.129*0.661/0.0017 = 50.16 Watts

Since roughly 745 Watts = 1HP,

50.16 Watts = 0.0673 Horsepower


Seriously, I don't see the error in this logic (not saying there isn't any, just that it's not obvious to me).

SteveM.
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Steve, give me a minute and I'll put a description of why this is incorrect. I have just had a conversation with Vibe, and he is absolutely correct (at least for this discussion and we're not nit picking the details.). The error is not in the calculation of this particular part of the equation, it is in the fact that we are not taking into account ALL of the things that are happening. These things assume that the projectile is fired in a vacuum (no atmospheric losses).

Hypothetically speaking, this bullet WILL ONLY DECELERATE 32fps due to gravity in the first second. Now, at that time, the remaining energy imparted in the bullet from being fired, (ENERGY TO ACCELERATE THE BULLET TO OUR SPEED) IS STILL IN THERE.

Gravity is our only source of hypothetical loss. So most of the energy is still in the bullet getting carried to the next 3000 feet of elevation. In fact, a VERY large portion of it.

Vibes calculations show the bullet will travel a hypothetical 26+ miles straight up (ignoring atmospheric losses) and this is correct.

The time to impart the energy at our hypothetical barrel length which he approximated at 26" but is another difficult number to nail down, represents a lot of energy transfer in a very short amount of time.

Short answer without going into TONS of math is,

VIBE IS CORRECT. 4MESH IS WRONG:eek:
VIBE IS CORRECT. 4MESH IS WRONG:eek:
VIBE IS CORRECT. 4MESH IS WRONG:eek:
VIBE IS CORRECT. 4MESH IS WRONG:eek:
VIBE IS CORRECT. 4MESH IS WRONG:eek:
VIBE IS CORRECT. 4MESH IS WRONG:eek:
VIBE IS CORRECT. 4MESH IS WRONG:eek:
VIBE IS CORRECT. 4MESH IS WRONG:eek:
VIBE IS CORRECT. 4MESH IS WRONG:eek:
VIBE IS CORRECT. 4MESH IS WRONG:eek:
VIBE IS CORRECT. 4MESH IS WRONG:eek:

Sorry Vibe, I tried size=x inside the UBB tags but it won't make it bigger:D:D

Horsepower on this hypothetical cartridge is definitely somewhere north of 4000 HP!

Very good Job VIBE! And THANKS for TEACHING ME that energy does not equal work BUT BUT BUT BUT ENERGY IMPARTED DOES = WORK!

4Mesh looks at HUGE plate of raw crow in rather large pieces and prepares for long happy meal.

Steve, the only analogy I can give that will explain where the rest of the HP is, is that we didn't LIFT the bullet (as Vibe said was an incorrect term I beleive), we accelerated it. I could not understand why that was getting included as work when we only move the bullet 3000'. Moving the bullet 3000 feet up is one part of the work. Making it go 2968fps PAST that point is the rest.
 
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After re-reading the thread again


Vibe said:
OK. I see what you're asking I think...Give me a bit to work it out and make it clearer...but the main issue is in the difference in the "acceleration term"


the 20ft/sec example is a bit different. IF you are saying that a 70,000 grain mass is traveling at an average of 20ft/sec and goes 10 ft...then no work was done, as the velocity did not change. IF on the other hand you are saying that it started at 0 velocity, and traveled 10 ft under constant acceleration in 0.5 sec resulting in an average speed of 20 ft/sec...then the final speed is 40 ft/sec at the end of that time period V(ave)=(V2-V1)/2 then if V1 =0, then V2 has to equal 2V(ave)

10/32 Slug * 40 * 40 ft^2/Sec^2 = 0.3125 * 1600 Slug ft/sec^2 - ft = 500 ft-lbs work

500ft-lbs work done in 0.5 sec is 500/0.5 = 1000 ft-lb/sec power.

If you are saying that this happened in a straight up direction then the previous 100 ft-lb of work due to change in potential has to be added to the 500 ft-lb kinetic energy increase due to velocity.
SO total work in this particular case would be 600ft-lbs
done in 0.5 sec would be 1200 ft-lb/sec of power.

Sorry this took so long. I had a virus crash during posting and have been diagnosing that as I posted.
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And right here is where I completely missed the boat on this. This was explained to me and I blew right past it.
 
Whatever the exact HP calculated figure is, it will be in several thousands of HP.

Rad
 
Vibe, I'm still in denial about why this is coming out to be true when some of these figures are what they are. First, the .03 pound bullet, and I guess what I consider a reasonably low velocity...

In attempting to apply some common sense to this, I thought about how much fuel is consumed on the dragster vs bullet comparisons.

In a reference way above, a dragster uses 1.5 gallons of fuel per second.

Our Win Mag uses about 75 grains of 4831 to make 3k fps. If the .0017 Sec follows, and actually there were estimates somewhere of .0012, those two values could mean we are burning powder at a rate of between 6.3 and 8.9 pounds of powder per second.

Now this kinda shows a real world example that someone not doing the math could look at and determine the winner, if you will!

Nice going again. And thanks again.
 
Another twist in the problem or the worm turns.

Vibe said:
210 grain * 1 pound/7000 grain * 1 slug/32 pounds = 0.0009375 Slug
(3000ft/sec)^2=9000000 ft^2/Sec^2

1/2MV^2= 1/2*0.0009375 *9000000 Slug-ft/sec^2 -ft

1/2MV^2= 4218.75 ft-lbs energy

Energy in chamber = 0
Energy at muzzle = 4218.75

(Energy at muzzle)-(Energy in chamber) = 4218.75 ft-lbs = work done

How much time did it take to do the work?
Again I'm going to assume a 26" barrel
Velocity at muzzle = 3000ft/sec
Velocity in chamber = 0
Assuming constant acceleration V(ave)=(Vmuzzle-Vchamber)/2=1500ft/sec
distance =Vt
so
T=d/V =(26" * 1ft/12")/(1500ft/sec)=2.166666667/1500=0.001444444 seconds

Work done/time to do the work= 4218.75/0.001444444=2920673.077 ft-lbs/sec

Work done in terms of HP
(2920673.077 ft-lbs/sec)/550ft-lbs/sec=5310.314685HP


It was a bit low. The 0.001711 sec. barrel time get's it a bit closer though, but I don't really know where that came from.
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Vibe,

Using your method, I have another problem for you to consider:

A 210 grain bullet strikes an immovable block of hardened steel at 3000 feet per second and makes a cavity 1/4" deep and drops to the ground. How much horsepower is required to slow the bullet to a full stop in 1/4"?

Concho Bill
 
552273HP unless I'm wrong again.

Which is entirely possible.

Actually Bill, I discussed this very thing while on the phone with Vibe today. I was concerned about my earlier supposition that the barrel length had nothing to do with the calculation. That was not a correct supposition. Barrel length means a lot and that's why he was estimating it.

Unless I've learned nothing, or somehow there's more to the story, HP would be inversely proportional to the barrel length and time used to generate the energy, or the distance and time used to stop it. Yes?
 
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OK OK, you'se guys have surpassed me with the math........I still need a couple of clarifications here :)


So in other words if one were to somehow gear down a 300WinMag it would really lift 550lbX4000 or 2,220,000lb one foot in one second? ONE 300WinMag round?

A 300WinMag round generates enough energy to lift over 2 million pounds over the course of one second?



Put it in simple terms for me.......am I WRONG too?????


Gearing up for my dish of crow here :D:D:D


al



BTW, Lynn, do you realize that your posts aren't showing up on the board?
 
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