westbronco

If you neglect air drag, variation of air density and temp, why bother? The calculation is meaningless.

Bigbuckslayer3

This is way over my head.

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jjt

i just come in from shooting the 8g .308 bullet 800 m/s straight up and measured the height and time they are as follows

height 10687 ft 7 7/16 in
time to reach elevation 37.31 seconds
and 74.82 seconds to fall back to the ground

however this took place at 6500 ft starting elevation so it will be different at sea level

driftrider

Well, I'm man enough to admit when I'm wrong, and well, I'm wrong. I made the mistake of using -9.8m/s^2 for the accelleration coefficient in the position function rather than the appropriate -4.9m/s^2. The reason the coefficient is -4.9 is because the position function is the anti-derivative of the velocity function.

Bigcountry had it right from the start.

Here's how the calculus breaks down to arrive at the solution, if you're curious as to WHY the functions are as they are, for those who are interested, rather than the usual process of finding an appropriate equation and plug in the numbers. I personally am the type that likes to know the "why" along with the "how".

So here goes:

First we start by knowing that the acceleration of gravity ( which we'll call a(t), with acceleration given as a function of time 't') is constant, and the value of the constant is 9.8m/s^2. Because the acceleration is slowing the bullet down initially we make the acceleration negative. Hence:

a(t)= -9.8

With calculus we know that the acceleration function of straight line motion is the derivative of the velocity function [v(t)], and v(t) is the derivative of the position function s(t). Since we only know a(t) at this point we muct work backward with calculus and find the anti-derivative of a(t), which is v(t), and then find the anti-derivative of v(t) which is the position function h(t).

So the second step is to find v(t) by finding the general anti-derivative of a(t) and using what we know about the situation to find the exact function. Using the rules for anti-derivatives we find that, in the general form:

v(t)=A(t)= -9.8t + C where C is an arbitrary constant. In this case, C represents the initial velocity of the bullet leaving the gun, which is given at 800m/s.

Therefore v(t)= -9.8t + 800

To find the position function, s(t), be again find the general form of the anti-derivative of v(t) which is:

h(t)=V(t)= (-9.8t^2)/2 + 800t + C where C is the initial position of the bullet at t=0 seconds. We can assume that the bullet is fired from ground, and since we aren't factoring in atmospheric effects (i.e. air density, temp, drag, etc...) then we're only interested in the vertical height from the shooters reference, which is zero meters height.

Therefore: h(t)= -4.9t^2 + 800t + 0 or to simplify and factor, h(t)= -t(4.9t - 800)

To know exactly where the local (and in this case absolute) minimum of the height function we simply set h'(t)=v(t)=0 and solve for t.

v(t)= -9.8t + 800 = 0 therefore t=81.6 seconds to apogee. Plugging 81.6 into the height function gives us a value, at apogee, of 32,653 meters (rounded to the nearest meter)

Finding the zeros of the height function tells us the time (t) when the bullet is at the level of the shooters muzzle (which we call zero height), which is t=0 and t=163.2 seconds.

And now you know the rest of the story!

Mike

eldeguello

From the BULLETS FROM THE SKY chapter of Hatcher's Notebook: "From calculations and experiments, the British concluded that their Mark VII bullet" (174-grain spitzer boattail @ a MV of 2440 FPS) "when fired vertically rises to about 9,000 FEET, taking 19 seconds to go up, and 36 to come down, total time of flight up and back 55 seconds. The Germans give similar figures for 7.92mm, their 153-grain flat-base spitzer bullet having a muzzle velocity of 2882 FPS. It is supposed to rise 9000 feet, taking 19 seconds to go up and 38 seconds to come down."


I submit that the 9000 feet max altitude Hatcher gives is much closer to what a 7.62mm NATO bullet would attain, (probably somewhat less for the 147-grain flatbased .308" NATO bullet), rather than the figures given above that show such outlandish predictions as 32,653 METERS!! It might, on the MOON! But you CANNOT IGNORE air resistance if you are shooting on a planet which has an atmosphere! Hell, even the 120mm antiaircraft gun has a vertical range of only 12 miles (63,360 FEET) with charge 8!!

driftrider

The calculations are accurate if we, like all of us specified, disregard drag. I think everyone here also realizes that drag is a major factor and knew that the 32,000m altitude was little more than a curiousity than an accurate prediction of the bullets actual attained altitude. None of use here are ballisticians and therefore know the very conplex formulas and calculations necessary to predict a bullets velocity when fired vertically, perticularly since the pressure, temperature and density of the air varies significantly in the vertical plane. We also didn't have the book you sited either. Because of this these predictions were acknowledged from the beginning as little more than for curiousity and amusement, and not meant as an accurate predictor of actual events. Of course, if I were to be nitpicky I could make the claim that your informations predicted altitude of 9,000ft is not accurate or applicable to the .308", 147gr NATO ball bullet because it has a different weight, diameter, BC and muzzle velocity than either the English or German bullets your data was based on and is itself meaningless.

Judging by your liberal use of exclamation marks (!) here and in all your posts I'd guess that you are very high strung, as you seem to exclaim everything.

FYI: If the same bullet were fired on the moon where atmosphere is negligable and really can be ignored, the bullet would reach an altitude of roughly 200km in 500 seconds. The average gravitational accelleration of the moon at the equator is 1.6m/s^2, which is almost exactly 1/6 of Earth's gravity.

Mike

eldeguello

[:-]Do you think that pointing out the discrepancy between 32,653 METERS and 9000 FEET is being nit-picky?? Wow

Yes, you are right that the NATO ball round will not perform EXACTLY the same as the 7.92 German, or the .303 Mk. VII. However, I would be willing to bet that it would not be more than 5% off one way or the other. Just look how closely the performance of the flatbased 153-grain 7.92mm (SD .2095) at a MV of 2880 FPS and the 174-grain boattail .311" bullet (SD .260) at a MV of 2440 FPS resemble each other, despite the fact that their ballistic coefficients and muzzle velocities are quite different. I suggest my prediction for maximum altitude for the .308 is not nearly as meaningless as yours.

Here you're beginning to sound like Fasier Crane. If you are a psychologist, you should stay out of ballistics discussions.

Now correct me if I misread John Bear's inquiry, but I interpreted what he said to be a question about how high the .308 bullet would go, ON EARTH FOR REAL, not somewhere where there is no atmosphere. He did not seem to me to be asking a an Alice-in -Wonderland question, but the answer he got was anything but reality.

(PS: If you don't have a copy of HATCHER'S NOTEBOOK, you should get one and read it. It might just prevent you from having to re-invent the wheel each time someone asks you a question to which the answers are already known. Maybe.

driftrider

Don't have to be a shrink to see that you're high strung...or perhaps overreactive is a better word.

As for John's question, what you take as being a dead serious question warranting a serious and accurate answer, I took as mearly a question posed out of idle curiousity. But thank's for correcting us, oh revered holder of the sacred Hatcher's Notebook! I'm sure that we'll all sleep much better tonight knowing that a .308 won't shoot higher than about 9,000ft, and I'm be sure to remember this very worthwhile tidbit of trivia and allow Santa to decend below 9,000ft before I start taking potshots at his reindeer with a .308win. Or better yet, I could just go with a .300WinMag. That should give me enough power to take out Rudolph at at least 10,000ft. The little sucker will be in my freezer before he knows what hit him! [8D]

Dude, lighten up before you bust an artery in your head!

Mike

eldeguello

Dude, when one runs out of something meaningful to say, one often resorts to getting personal. Now, have I made any disparaging remarks about YOU?? BTW, nothing concerns me enough to cause me to bust an artery!