Math question re bullet drop

[carlsbad]

P.S. Since I gave the correct answer to the OP's question 33 posts ago, it's not clear why this thread continues.

You didn't give an answer. The answer would be a number. See my (correct) answer. --Jerry

 

[carlsbad]

For those of you arguing acceleration of g, you are all somewhat correct but generally do not know why. indeed the bullet is at all times, while near the surface of the earth, undergoing a gravitational acceleration of g. Until you fire it, it is held up by the gun.

After you fire it, it is un-restrained and is accelerated by gravity of magnitude g, in the direction of the center of the earth. It is also acted on by other forces and has an initial velocity.

To define the flight time, final velocity in z direction, etc you just have to add up all the z velocity and various accelerations that act upon it. saying you can ignore all but gravity is incorrect.

--Jerry

 

[hogpatrol]

Way overdue.

 

[M-61]

Way overdue.

Maybe it is.....BUT it beats the hell out reading which bore cleaner is best! Some really interesting points I thought. I can't do the 'how to break in a barrel' anymore.

My thanks to all!!

 

[Toby Bradshaw]

You didn't give an answer. The answer would be a number. See my (correct) answer. --Jerry

Hi Jerry. You did see the "quarter inch" in my first post in this thread, right? And since the OP never specified the parameters of the cartridge in question, your answer isn't necessarily "right," either.

OK, got a manuscript to write. Y'all carry on without me.

 

[dickn52]

To prove whether a bullet can enter and pass thru a 1½ "dia. tube, one foot long, without touching the inside of the tube from 450 yards.
From sheepdog's post #21 it looks like there is room to spare. Bit more than ¼" in one yard of drop if I'm following it correctly

Oh, the old shoot the sniper through the scope scenario.

 

[carlsbad]

Hi Jerry. You did see the "quarter inch" in my first post in this thread, right? And since the OP never specified the parameters of the cartridge in question, your answer isn't necessarily "right," either.

OK, got a manuscript to write. Y'all carry on without me.

No, I didn't see the "quarter inch". I was looking for a number. Is that enough significant digits to satisfy the OP?

A complete answer to a physics problem always has to specify the parameters unless they are obvious. I even specified the altitude (the only affect of altitude, if the atmospheric conditions are specified elsewhere, would be a variation in the value of g).

--Jerry

 

[Brians356]

No, I didn't see the "quarter inch". I was looking for a number. Is that enough significant digits to satisfy the OP?

A "quarter inch" = 0.250000... inches. Is that enough significant digits to satisfy the gadfly?
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[carlsbad]

A "quarter inch" = 0.250000... inches. Is that enough significant digits to satisfy the gadfly?
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In mathematics, one would assume 1/4" is +/- 1/8". a gadfly is about 1/4" long so 1/8" would be a big deal to them. --Jerry

 

[Brians356]

In mathematics, one would assume 1/4" is +/- 1/8".

Really? They must have adopted that rule sometime after I earned my mathematics degree thirty years ago. Good to know, thanks!
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[DRNewcomb]

Most posters talk about the acceleration of gravity but ignore the fact that the slightly nose-up orientation of the bullet imparts a small but significant lift as it flies through the air.

 

[Alexander-M]

I hadn't tried the Berger Ballistic Calculator because I didn't think it would have 1-yard increments, but I was wrong.
Here are the results: With a zero at 450 yards, it drops 0.17 inches at 451 yards.


If I expand the range to 420 - 480 to get a better idea of the time-of-flight (TOF) per yard, it shows that it takes about six to seven yards in this yardage span for a 0.1 second TOF change, or approximately 0.014 seconds to travel one yard. The formula for a free fall object "in a vacuum" I used earlier gives an answer that is way off. I tried it again with different parameters, and it is obvious that the air resistance is rather significant.

As I just mentioned, the ballistic calculator yields a time of approximately 0.014 seconds for the bullet to travel in this span. But if try to calculate this time using the speed of the bullet at 450 yards; i.e., 1934 fps per the chart, which equals 644.7 yards/s, we would get 0.00155 seconds and not 0.014 seconds. Maybe after the BSWN Bryan could chime in.

Alex

 

[carlsbad]

Most posters talk about the acceleration of gravity but ignore the fact that the slightly nose-up orientation of the bullet imparts a small but significant lift as it flies through the air.

Theoretically, from an aerodynamic perspective, the bullet should have no lift. Unlike an airplane wing, the bullet is symmetrical from top to bottom. An airplane wing gets lift by the Bernoulli principle when air flowing over the top of the wing has a higher relative velocity, thus a lower pressure. This shouldn't happen in a bullet so simple physics says there is no lift. Now second order effects might provide minimal lift and this is somewhat controversial. Spin drift might raise a bullet a little. Either way, the lift is minimal to none.

More importantly, we should remember that even a rifle zeroed at 100 yards is aimed slightly upward. So even at point blank range, a bullet has some upward velocity when it leaves the muzzle.

--Jerry

 

[carlsbad]

Alex's results for a 168 grain bullet show that it is doing much better than a 155 grain Amax at 450 yards. Although expected because the 168 berger has a much higher bc, I'm a bit surprised by how much it it is different. This is a great demonstration of the advantages of high bc bullets at mid to long ranges. --Jerry

 

['Freak]

What’d be the angle of the tube, shooter’s end up from horizontal, to have the perfect shot drop down through its center ...?

 

[DRNewcomb]

Theoretically, from an aerodynamic perspective, the bullet should have no lift. Unlike an airplane wing, the bullet is symmetrical from top to bottom. ....

This would be true if the bullet's orientation was always parallel to the direction of flight. However, at longer ranges, this is not the case. Spin holds the bullet in the same orientation it had when it left the barrel (i.e. slightly upward). This divides the drag vector into two components, one horizontal and another vertical. The vertical component is also called "lift". Various texts deal with the phenomenon in different ways. The Sierra Reloading Manual (ed. 2, p. 379) "As the bullet path curves downward as the range increases, the drag vector tips upward and a small component of drag then begins to act along the vertical direction, parallel to gravity." Hatcher deals with this on p. 626, under the guise of air resistance slowing the rate of drop. I've read and forgotten so much exterior ballistics I can't recall where I saw it actually dealt with as "lift".

 

[carlsbad]

I agree 100% that that the orientation would stay as it came out of the barrel. 250,000 rpm is a very powerful gyroscopic stabilizer. So given that the "flying brick" space shuttle has enough lift to get back to earth, I'll buy this explanation. --Jerry

 

[expiper]

why doesnt someone take a hi speed slo/mo camera out to the target and take a pic of the bullet coming in to the target at 1000 yds or so,,..(If I had one I shure would ),,we could see the bullet comein in to the target like a plane on aproach and watch it pass thru the target,,this will prove who knows what they are talking about when it comes to ballistics,,!!!...Roger