Bullet dropped lands the same time as a bullet fired..

[lysanderxiii]

Nope in this case you would still need a good estimate of the BC or a velocity point.

IE:
A bullet with a .5 BC (a good 6.5mm - 30 cal bullet might have a BC like this) launched at 2000 fps from a perfectly level barrel would drop 4.91 inches (using a G1 ballistic model) at 100 yards
A bullet with a .095 BC (a light 30 cal pistol bullet might have a BC like this) launched at 2250 fps from a perfectly level barrel would drop 4.91 inches (using a G1 ballistic model) at 100 yards
A bullet with a .0572 BC (a 410 slug might have a BC like this) launched at 2500 fps from a perfectly level barrel would drop 4.91 inches (using a G1 ballistic model) at 100 yards

No doubt I can calculate any muzzle velocity I want (with in reason) simply by tweaking the BC to get me roughly a 4.91 inch drop.

You need two velocity points to calculate the BC. Assuming you know the velocity at the muzzle, and can calculate the velocity at some range due to drop, you should be able to get the AVERAGE BC over that speed range.

 

[mcb]

You need two velocity points to calculate the BC. Assuming you know the velocity at the muzzle, and can calculate the velocity at some range due to drop, you should be able to get the AVERAGE BC over that speed range.

Yes velocity at two (or more) points down range can certainly be used to estimate a BC if you assume a particular ballistic model.

That said if you know the muzzle velocity and the drop at a know range (a measurement that would require some averaging over many shots to eliminate the inheritance precision (or lack there of) of the weapon firing the bullet) you could also make a reasonable estimate of the BC, again assuming a particular ballistic model. Within a give ballistic model (ie G1, G7 etc) only one BC will give you X drop at Y range with Z muzzle velocity.

 

[ATLDave]

^ Agreed. That's because drop is determined by velocity, and decay of velocity is what BC is for! If you know the velocity at one point and the subsequent drop, that's a good way to calculate BC. Or for someone other than me to calculate BC. I was told there would be no math!

 

[stchman]

Simple Physics.

 

[RecoilRob]

Seeing as this conversation has drifted a bit to the esoteric side of physics....let me mention that two objects of differing mass dropped simultaneously in a vacuum do NOT reach the ground at the same time. The one with more mass will drop faster...if only by a tiny amount. The word 'same' is only as valid as your ability to measure is. Two rods measured with a tape will look to be the 'same' if within 1/64" or so.....but easily seen to be different with a more precise measuring device. So it goes with the gravity drop test.....measure close enough and the heavier object will win the race to the ground by a tiny fraction.

 

[mcb]

Seeing as this conversation has drifted a bit to the esoteric side of physics....let me mention that two objects of differing mass dropped simultaneously in a vacuum do NOT reach the ground at the same time. The one with more mass will drop faster...if only by a tiny amount. The word 'same' is only as valid as your ability to measure is. Two rods measured with a tape will look to be the 'same' if within 1/64" or so.....but easily seen to be different with a more precise measuring device. So it goes with the gravity drop test.....measure close enough and the heavier object will win the race to the ground by a tiny fraction.

I don't believe that is correct, in a vacuum both the light and heavy objects will accelerate at exactly the same rate and thus fall at the same velocity assuming they are released at the same time from the same altitude. That math is relatively easy to show.

Gravitational Force F = G M m / r^2

G = Gravitational Constant (fix value across the universe)
M = Mass of the earth (or any planet or object we want to be attracted to)
m = mass of object we are going to release
r = Distance between the centers of mass of the M and m

Fundamental equation of motion (ie Newton's second law of motion). This equation governs the motion of all objects and is accurate for nearly all situations (moving at a large fraction of the speed of light breaks it, so does close proximity to black holes, and a few other situation but otherwise its good to go.)

F = m a

F = Force
m = Mass (in this case the mass of the object we will drop)
a = Acceleration

Set these two equation equal to each other. Which is basically asking what will the m object's motion be like under the force of gravity.

m a = G M m / r^2

Algebra tells us we can cancel the m's on both side of the equation. This then leaves us with the acceleration due to gravity

a = G M / r^2

Thus the mass of the m object does not factor into the acceleration the m object experiences toward the M object. ie a heavier m object produces a proportionally larger force thus maintaining the same acceleration. All m objects at the same radius from M will experience the exactly the same acceleration independent of their own mass. Barring other forces at work both object should experience the exact same acceleration and fall exactly together.

G = 6.67408 × 10^-11 m3/kg-s^2
M = 5.972 × 10^24 kg (for the earth
r = 6.371 x 10^6 m

a = 9.820 m/sec^2 (this is an average value since we used an average radius. The earth is not a perfect sphere. This value is slight higher than the usual measured value due to not accounting for the centrifugal acceleration most of the surface experiences due to the planet's rotation, the poles zero, max as the equator. Science often uses the average measured value 9.80655 m/s^2

Rambling...

 

[RecoilRob]

That always generates a good response. I DO think that many mathematical equations are made to simplify things by the people making them and to many engineers...these are gospel and you are a heretic if you challenge them. For example: a senior engineer at Pratt & Whitney once told me that it was impossible to run the quarter mile in less than (I forget his numbers...maybe 5 seconds) and was in total disbelief when I told him that dragsters regularly ran in the 4's at that time. He said "IMPOSSIBLE"! and proceeded to fill half a blackboard with equations to prove he was right. "But....they do it every weekend...maybe you should tell them it's impossible". Man...did he get mad... but obviously he was missing something in his equations...right?

Isn't it possible that the equations you quoted are simplified just because the actual differences really muck things up and for the most part don't matter in the end? We're talking about really small changes here...but when theory is involved I always try to be aware when rounding or simplification is involved. You might not remember when the rocket guys first tried hitting the moon...do you? Well...if you don't, they missed. Bad. Several times...and their equations couldn't figure out why. So they (at first) just measured how far they had missed and then added a correction factor to actually hit the darn thing. I never found out exactly WHY their equations were wrong enough to miss the moon, but I do suspect that the acceleration of gravity constants you quoted were part of the reason. Yes...they're very close and don't matter for most things...but in this case those little rounding errors were huge.

OK. Practical exam time!! This question was on my A&P Practical Exam and I got it wrong. You have a blimp flying at 5000 ft, at the equator, at high noon on a clear sunny day. It casts a shadow on the ground. Is that shadow (a) larger than the blimp. (b) smaller than the blimp. or (c) the same size? I'll not ruin the suspense by answering it here and see what you have to say.

 

[GAF]

A Bowling Ball And Feather Will Fall At The Same Rate In A Vacuum Chamber. ... If you simultaneously drop a bowling ball and a feather from the same height in a vacuum, they will hit the ground at the same time.

 

[ATLDave]

Seeing as this conversation has drifted a bit to the esoteric side of physics....let me mention that two objects of differing mass dropped simultaneously in a vacuum do NOT reach the ground at the same time. The one with more mass will drop faster...if only by a tiny amount.

I don't think that's correct, and it's neither because of oversimplified equations nor measurement resolution limitations. You seem to overlooking the fact that the falling object has inertia, which is resistance to acceleration. The heavier object may be contributing its own mass to the gravitational forces, but those contributions are precisely and directly offset by the greater inertia of the object which resists the downward acceleration. That's why the same variable is in both sides of the equation. That's not a math trick that mcb is pulling.

But I'm just a non-physicist/non-engineer dude on the interwebs, so I could be wrong.

 

[mcb]

That always generates a good response. I DO think that many mathematical equations are made to simplify things by the people making them and to many engineers...these are gospel and you are a heretic if you challenge them. For example: a senior engineer at Pratt & Whitney once told me that it was impossible to run the quarter mile in less than (I forget his numbers...maybe 5 seconds) and was in total disbelief when I told him that dragsters regularly ran in the 4's at that time. He said "IMPOSSIBLE"! and proceeded to fill half a blackboard with equations to prove he was right. "But....they do it every weekend...maybe you should tell them it's impossible". Man...did he get mad... but obviously he was missing something in his equations...right?

Isn't it possible that the equations you quoted are simplified just because the actual differences really muck things up and for the most part don't matter in the end? We're talking about really small changes here...but when theory is involved I always try to be aware when rounding or simplification is involved. You might not remember when the rocket guys first tried hitting the moon...do you? Well...if you don't, they missed. Bad. Several times...and their equations couldn't figure out why. So they (at first) just measured how far they had missed and then added a correction factor to actually hit the darn thing. I never found out exactly WHY their equations were wrong enough to miss the moon, but I do suspect that the acceleration of gravity constants you quoted were part of the reason. Yes...they're very close and don't matter for most things...but in this case those little rounding errors were huge.

The rambling from my previous post were all couch in classic Newtonian physics and are as correct as far as we have ever observed. They are no simplifications or approximations in those equations within that model. That said we could set the problem up in Einstein's General Relativity (the current accepted and best model of gravity) but we would be stepping outside of my formal training as I only made it to Special Relativity in my formal physics classes. That said I have studied General Relatively just enough to know that even if we break out that hard math (how are your tensor calculus skills? Mine are very rusty...) that General Relativity would predict both objects, the heavy and light, would move exactly the same in the warped space-time created by the mass of the earth assuming that it is the only force acting on both objects. ie in this particular example of dropping both a bowling ball and a feather in a vacuum General Relatively would reduce to the same results as the Newtonian model.

OK. Practical exam time!! This question was on my A&P Practical Exam and I got it wrong. You have a blimp flying at 5000 ft, at the equator, at high noon on a clear sunny day. It casts a shadow on the ground. Is that shadow (a) larger than the blimp. (b) smaller than the blimp. or (c) the same size? I'll not ruin the suspense by answering it here and see what you have to say.

Without thinking to hard about it to hard I would expect the shadow would be both smaller and larger assuming the blimp was large enough to completely occlude the sun at that altitude. Think of it like a solar eclipse. The umbra (the darkest part of the shadow) would be smaller and the penumbra that partial shadow cast would be larger than the blimp. If the blimp was too small to occlude the sun then the penumbra get even fainter and the umbra gets replaces by a faint antumbra (ie you see the sun as a ring round the blimp).

 

[RecoilRob]

Well....you can't say both (a) and (b)...so you fail. I got it wrong because I said (a) larger than because the 'rule of shadows' says that a shadow is always larger than the object that created it. They said the correct answer was (c) the same size...even though it is technically wrong. Their logic was that the sun being 93,000,000 miles from earth and the blimp being only about 1 mile above the ground...for all intents and purposes the shadow would be the same size. I countered that if they'd have put 'for all intents and purposes' in the question I would have agreed that it would be the same...but 'technically' it MUST be bigger...and in the end they gave me credit for that answer.

The whole deal about the heavier object falling faster and the general teachings about it came from a fellow in the Rocket Dept. at Pratt & Whitney during a conversation one evening. IIRC he said that 'for most intents and purposes' the equations that everyone is taught work just fine...but 'technically' there is a bit of 'fudge factor' there to simplify the equation. How would it work if one of the 'm' values instead of being equal and cancelling was .0000001 larger or smaller? In most cases it would do nothing but muck up the calculations by extending each number out to a ridiculous extent...so it is dropped. This may not have even been known until they found their calculations couldn't hit the broad side of the moon. I can't say that I'm a math expert or can even get my head around the concept of gravity...I just enjoy a good conversation and reading the thoughts of intelligent people when the chance presents itself. Thank you for your reply.

 

[ATLDave]

It's not just unnecessary to use relativity based calculations (which are necessary for precise calculations of things on the scale of space flight) for small-scale measurements, it can get to the point where you run into the as-yet-unresolved conflict between relativity and quantum mechanics. If your relativity calculations show that the heavier object is beating the lighter object by less than a planck length....

 

[mcb]

Well....you can't say both (a) and (b)...so you fail. I got it wrong because I said (a) larger than because the 'rule of shadows' says that a shadow is always larger than the object that created it. They said the correct answer was (c) the same size...even though it is technically wrong. Their logic was that the sun being 93,000,000 miles from earth and the blimp being only about 1 mile above the ground...for all intents and purposes the shadow would be the same size. I countered that if they'd have put 'for all intents and purposes' in the question I would have agreed that it would be the same...but 'technically' it MUST be bigger...and in the end they gave me credit for that answer.

The whole deal about the heavier object falling faster and the general teachings about it came from a fellow in the Rocket Dept. at Pratt & Whitney during a conversation one evening. IIRC he said that 'for most intents and purposes' the equations that everyone is taught work just fine...but 'technically' there is a bit of 'fudge factor' there to simplify the equation. How would it work if one of the 'm' values instead of being equal and cancelling was .0000001 larger or smaller? In most cases it would do nothing but muck up the calculations by extending each number out to a ridiculous extent...so it is dropped. This may not have even been known until they found their calculations couldn't hit the broad side of the moon. I can't say that I'm a math expert or can even get my head around the concept of gravity...I just enjoy a good conversation and reading the thoughts of intelligent people when the chance presents itself. Thank you for your reply.

Not sure what missions you are talking about missing the moon? All the lunar mission I have ever seen that failed to get to the moon missed because of technical problems with the space craft not due to miss calculated orbital parameters. We had General Relativity 42 years before the first attempt to launch a rocket to the moon. We have never missed the moon due to a miss understanding of orbital mechanics, lots of other reason but not that.

 

[GAF]

A bullet fired from three feet height And a Feather dropped from three feet hieght at the same time as a bullet is fired Will Fall At The Same Rate In A Vacuum Chamber. They will hit the ground at the same time.