Physics question

"Dean A. Scott" wrote in message
news:1126146846.d1f75da4b1eff1487f40a17ade08c409@t eranews...
Here's a wacky scenario... why can you fire a gun on the equator
in the direction of the setting sun? I mean, the Earth spins at
the equator at around 1,000 miles per hour (24,000 mile circum-
ference divided by 24 hr rotation), so why does the bullet exit
the barrel at the same velocity no matter what direction you
point it in? Hmmm. :-)

Imagine that the target is going away from you at the same speed that the
bullet is traveling. It would hit the ground before it reaches the target.
(Actually, it would never catch the target.

Now, turn that around. Imagine the target is standing still and you are
going away from it at bullet speed. Same ting happens, mon. De bullet hits
de ground.

Like incest, it's all relative.

Rich S.

"Montblack" wrote in message
...
wrote)
Won't it drop stairght down toward the manhole cover? (Yes it will.)


Gads, you're right! Crap, now it's basically T-ball at the manhole cover.
There goes my 1.35 ERA!!


Montblack

I had a .98 GPA one semester. Only courses I passed were Physics 101 and
ROTC.
Nobody flunked ROTC.

Oh! ERA. Never mind.

Rich S.

"Harry K" wrote in message
oups.com...

The way I see it, you are both in agreement. Fred says to hit the 109
he has to fire at the exact time the 17 croses the 109's path. That is
correct. In effect he says that to fire as per your example he can't
hit it.

We are. Just had to define a few terms and conditions.

Rich S.

"Dan, U.S. Air Force, retired" wrote in message
news:cdOTe.9805$dm.6970@lakeread03...

Forgetting turbulence following the B-17 the bullet would still have a
muzzle velocity of 900 FPS, assuming a military load. 50 yards is 50 yards
and the Me109 would still be hit assuming proper aim. Whether it does much
more than penetrate the canopy I don't know. Put the trailing turbulence
back into the scenario and all bets are off.

Remember, the ME-109 is not following the B-17. It is crossing behind the
B-17 at 90° and is fifty yards behind the bomber at the moment their flight
paths cross. The bullet comes out of the muzzle with a velocity of 820 fps
relative to the pistol. The pistol however, is traveling 820 fps in the
opposite direction relative to the ME-109. Therefore the bullet has zero
velocity toward the ME-109 and drops straight down due to gravity.

Rich S.

"Dan, U.S. Air Force, retired" wrote in message
news:MjOTe.9806$dm.4606@lakeread03...
Blueskies wrote:

But just imagine, as the bullet falls to earth, it will remain horizontal
for a while due to the gyroscopic spin caused by the rifling...


No, the bullet will start dropping due to gravity immediately upon
exiting the muzzle.

But, it will drop in a horizontal attitude. Pretty much, that is. Since it
begins to accelerate downward, the axis of rotation will diverge due to
gyroscopic precession. Don't ask me in which direction, though.

Rich S.

" Blueskies" wrote in message
...
But just imagine, as the bullet falls to earth, it will remain horizontal
for a while due to the gyroscopic spin caused by the rifling...

Quite a while, actually. If you fire it straight up, it will likely land on
it's butt - still spinning. Ever done the math to figure the rpm of a .45
ACP? The twist of the barrel IIRC, is one turn in 9 inches. If it's
traveling at 820 fps at the muzzle, that is 1093 nine inch increments per
second or 65,580 rpm. No wonder the gun twists in your hand during recoil.

Rich S.

Rich S. wrote:
wrote in message
ups.com...

You're still assuming he's leading the ME-109 as if the gunner
were in a fixed location. He's assuming the gunner is actually
trying to hit the plane, and smart enough to figure out how
so that he times his shot *correctly*. You're both assuming
he fires straight back. He can hit the Me-109 by firing straight
back, but ONLY if he fires exactly as B-17 crosses the flight
path of the Me-109, and only if the Me-109 is fast enough or
close enough to get there before the bullet drops out of the
flight path of the Me-109. If the Me-109 is at a range of 50
yards, and no slower than the bullet the bullet will be no
more than 7 inches lower than the muzzle of the gun and will
still hit the Me-109 assuming the two planes are at the same
altitude.

Fred...............

I'll say it again - you misunderstand the question.

NO. I understood the question(s) and I answered them. The gunner
cannot hit the ME 109 by aiming and firing the ways you
described. There are an infinite number of ways the gunner
can aim and fire and miss, you picked some of them.

You also stated multiple versions with different
circumstances, not just different wording. It is not clear
that you realize that.

After answering YOUR questions, we then addressed the question
of how the gunner could hit the ME 109. That question was
NOT your question. That is a different question and I
don't see how I can make THAT any clearer to you.

No rule of nettiquette requires that we limit our discussion
to only the case you present. You are not the boss of us!
Courtesy requires that we make it clear when and how we
change the statement of the problem. We have made that
clear. You just don't realize that. We changed the
problem by relaxing the constraint on when the gunner
leaving him free to fire whenever he wants to.

Why? Because we're Americans goddamnit we want him
to score a hit. What are you, German?

clearly as I am able and even drawn a diagram. I don't know how else to
explain it. I'm a pilot. I talk with my hands (hopefully without resorting
to certain gestures).

The B-17 is flying West. The ME-109 is flying South. They are both at the
same altitude. There is no wind. The ME-109 will cross the flight path of
the B-17 and the B-17 will, at that moment, be fifty yards from the ME-109.
The tail gunner fires his weapon straight behind the B-17.

In *this* statement of the question, unlike
your first, you do not say WHEN he fires.
Since I WANT him to hit, I'll tell YOU
when he fires.

Got it!

He can hit the ME-109 by firing when the B17 crosses the flight
path of the ME-109. Then the Me-108 is due North of the gunner.
If he waits any longer, the Me-109 passes east of the bullet.
If he fires sooner, the Me-109 passes West of the bullet.

If he fires when the ME 109 is due North the Me-109 either
hits the bullet or passes over it depending on the range and
speed of the Me-109.

You specified that the Me-109 passes fifty yards behind
the B-17. A hit requires that the time
between firing and the time the ME-109 arrives behind the
B-17 is the time it takes the B-17 to travel 50 yards.
OP already calculated that, and also calculated that the
bullet drops 7 inches in that time. You did not
specify the speed of the ME-109 or the range when the
gunner fires, so we are free so vary those as needed.
there are several combinations of speed and range that
allow the ME-109 to be due North when he fires and
cross 50 yards behind. For instance, if the Me-109
is flying at the same speed as the B-17 and it crosses
50 yards behind it, then the Me-109 will also be 50 yards
when it is due North of the B-17 and the gunner can therefor
fire when the B17 is at a range of 50 yards and hit.
What a coincidence!

This time around I did not have to change the problem
because this time around you did not constrain the time
at which the gunner fires.



There is no way that he can hit the ME-109, given the stated speed of the
B-17 and the muzzle velocity of the bullet.

Wrong. The way is if he fires WHEN the B-17 crosses the flight path
of the Me-109. That drops a bullet in the flight path of the Me-109
and if the Me-109 is 50 yards due North when the gunner fires and
flying at 820 fps the Me-109 and the bullet are *both* 50 yards
behind the B-17 when the Me-109 crosses the flight path of the
B-17. The bullet has only dropped 7 inches, so it hits.


The ME-109 will *always* be fifty yards or more away from the bullet.
Always.


Not if the gunner fires when the B17 crosses the flight path of
the Me-109. That drops the bullet right in the flight path
of the Me-109 and the Me-109 closes on it as it drops.

Here is a hint. If you constrain all of the variables to specific
values the question is 'does', not 'can'. 'Can' implies at least
on variable is unconstrained as in your question above wherein
you left the time at shich the gunner fires free to vary.

--

FF

wrote in message
ups.com...

You also stated multiple versions with different
circumstances, not just different wording. It is not clear
that you realize that.

I did phrase the question poorly. I have said that *several* times and
attempted to clarify the wording - not change the intent of the original
question.

Now that that's over, I still find that it is an interesting problem in
relativity and deserved rational discussion.

Rich S.

"T o d d P a t t i s t" wrote in message
news
If you understood the physics, why did you bother to ask the question?

Is this not a forum for discussion? What may be "obvious" to you, may be of
interest to someone else.

Why do you bother to ask?

R i c h S .

"T o d d P a t t i s t" wrote in message
...
Because I had a real question - Why did you ask a question
you knew the answer to? Was it just because you thought it
was an interesting topic for discussion?

Bingo.

Rich