How’s my physics?

[PJ_in_FL]

Since the bullet's lateral or "X" velocity was the same as the rifle's when fired, but the bullet's altitude increased, it will "fall behind" the rifle as both circumnavigate the same point but at different altitudes.

Are you thinking of the Coriolis effect here?

Alan

The Coriolis effect is from movement across latitude lines where the motion is nearer or further than the equator. The demonstration would have shown no differences if:

• The demonstration had been performed correctly and the shots made in North and South directions
• The test was performed at the equator

The effect I described is simple rate calculations at different radii required to complete 360°s of motion in time t.

If the bullet's flight time is 10 seconds, reaching a maximum altitude of 1 mile, then the surface of the earth (gun) has moved 1.708 miles assuming a radius of 4000 miles. That's 9018.24 inches. I used the calculator here to do the heavy lifting.

Assuming an average height of .5 miles, the arc length at that height is 1.7082 miles or 9019.296 inches, which is the distance the bullet would NEED TO TRAVEL to keep up with the gun. Since the bullet moves the same distance as the gun (their linear "X" velocities are the same), it falls behind by the difference of the two, or 1.056 inches. Unless this is a blunderbuss, I don't think we'll see the bullet reenter the barrel when it falls back to earth.

The actual arc the bullet describes is much more complicated due to the changing velocity and height and would require considerably more maths than I'm going to attempt for this thought problem. Were I Issac Asimov writing one of his wonderful science essays that I ate up as a young lad, I might consider the effort. But for Chris, well ...

[ChrisGreaves]

The issue is that the ground rotates beneath the bullet. Your vertical shot is just at the extreme of a (idealized) parabolic trajectory. The principle is the same. Alan

Thanks Alan.
I see it now.
Also I have found some videos of ships sinking ....

[AlanMiller]

The Coriolis effect is from movement across latitude lines where the motion is nearer or further than the equator. The demonstration would have shown no differences if:

• The demonstration had been performed correctly and the shots made in North and South directions
• The test was performed at the equator

The effect I described is simple rate calculations at different radii required to complete 360°s of motion in time t.

The shots were performed "correctly". They were intended to demonstrate the Coriolis effect for the video, but shots like this are used by shooters and manufacturers to build data for moa corrections. That's what the knob on the top of the scope (in the video) compensates for.

If the bullet's flight time is 10 seconds, reaching a maximum altitude of 1 mile, then the surface of the earth (gun) has moved 1.708 miles assuming a radius of 4000 miles. That's 9018.24 inches. I used the calculator here to do the heavy lifting.

Assuming an average height of .5 miles, the arc length at that height is 1.7082 miles or 9019.296 inches, which is the distance the bullet would NEED TO TRAVEL to keep up with the gun. Since the bullet moves the same distance as the gun (their linear "X" velocities are the same), it falls behind by the difference of the two, or 1.056 inches. Unless this is a blunderbuss, I don't think we'll see the bullet reenter the barrel when it falls back to earth.

That sounds like a Coriolis-type correction to me.

The actual arc the bullet describes is much more complicated due to the changing velocity and height and would require considerably more maths than I'm going to attempt for this thought problem. Were I Issac Asimov writing one of his wonderful science essays that I ate up as a young lad, I might consider the effort. But for Chris, well ...

Indeed. And I think you'll find a lot more than simple maths is required too.
See, for instance, Design for Control of Projectile Flight Characteristics

Alan

[PJ_in_FL]

Coriolis "force" is shown by this image, which shows the effects of changing latitude in a rotating spherical reference frame. The "force" pushes things ahead if going to higher latitudes, and behind if going to lower latitudes.

x2scrap.png

What I described is essentially the bullet's reference frame is based on the point on the sphere it was at when it left the barrel of the gun. That reference frame is not accelerating but the gun's reference frame is, thus it's describing a curve around the axis of the sphere. Although both the bullet and the gun started out with the exact same horizontal component of speed (not velocity), that speed (scalar) is being applied in different reference frames so the end point location of the gun and the bullet will be different after a period of time.

A different effect is observed if you drop a ball bearing down the center of the Eiffel Tower. The ball bearing will land eastward of the center of the base because the top of the tower is moving faster eastward than the base (larger radii) and the ball bearing keeps the same eastward component of speed all the way down. The ball bearing's eastward component is higher than the base's eastward component, so it will move farther east than the base in the same amount of time even without taking the rotation of the base's reference frame into account.

In the first, the different reference frame accelerations cause the variation in end points, and the second, the different initial velocities cause the variation. The ball bearing example is closer to Coriolis effect than the bullet.

I'd go on but I need to go take an aspirin now ...

[AlanMiller]

I'm afraid I don't follow your "horizontal" argument. If we were doing this on the (rotating) moon, without the complications of atmospheric drag, then the fact that the moon continues to rotate beneath the projectile makes no difference. Horizontal velocity for both the moon and the bullet remain the same. The (0.5 mile) distance above the surface is trivial, and the distance covered by the "start point" on the moon's surface is so small that we can consider its surface to be a non-rotating, moving plane, as seen by an "inertial" observer in space.

Is this what you're thinking of?

Alan

[PJ_in_FL]

Ok, then let's use the Moon as our test bed.

It's rotational speed at the equator is 4.63 m/s, which coincidentally is just about 1% the speed at Earth's equator, but I digress.

Putting Chris's gun on the Moon's equator, let's shoot this bullet with LOTS of powder, so much so that the bullet goes to an altitude of 100 KM above the surface of the Moon. With the radius of the Moon being 1738 KM, then this altitude is 1838 KM from the center of the Moon.

Rise and fall times for that altitude are 351 seconds so let's make total flight time 700 seconds. Now, to simplify the equations (i.e. avoid messy things like integrals like the plague!), let's assume Chris has a magic gun and the bullet gets to that altitude in 0 seconds, but stays there the full flight time of 700 seconds then falls back to the surface in 0 seconds.

So with that mind problem set up, the surface of the Moon moves 3240 meters in those 700 seconds. That means that point on the Moon moves around the circle by 0.107 degrees. Now, the bullet, way up there 100 KM above the Moon moves the SAME DISTANCE since it's linear velocity was the same as the gun's. However, it's only traversed 0.101 degrees of distance around the Moon, so when it comes back down it will be 0.006 degrees behind the gun, or 176 meters in linear distance.

This scenario is a worst case distance since the flight time and altitude vary, but the bottom line is the gun goes around the Moon in a smaller arc than the bullet, so no matter what the height it travels it comes down behind the gun eventually.

[AlanMiller]

Ah, yes! Now I see where you're coming from. I think the fundamental problem with your analysis is that you haven't accounted for the fact that the direction of gravity is always radial, not vertical (as defined using your inertial reference frame). This means that the "residual" x-velocity of the projectile ("inherited" from the Earth's rotation) will be changing in magnitude continually throughout the trajectory.

I would try to solve this in the rotating reference frame of the Earth, but it would mean a refresher course of my Lagrangian mechanics, and I don't think there's enough aspirin in the house to take that on. Seems someone else has endured the neurological pain already though, (edit) using Newtonian mechanics in a non-inertial frame.

Alan

[ChrisGreaves]

... the direction of gravity is always radial, not vertical ...

Thank you Alan.
I had justed downloaded all three pages to re-read at home (I have Advil in the medicine cabinet) when I saw this.
The idea that we ought to be using radians and a radial view was beginning to flutter up in my brain.

As a primitive flatlander I tend to think of "up and down" relative to the "plane of the Earth's surface", but I am left with this nagging feeling that once the bullet leaves the rifle, it will be possessed by gravity and hence begin to describe some sort of orbital path.

(I'm beginning to wish I'd never asked .....)

P.S. If I do decide to take an Advil, i shall lie flat on my back and spit it out as best I can directly upwards and see if it plops back in my mouth!
Chris

[AlanMiller]

... the direction of gravity is always radial, not vertical ...

Thank you Alan.

I was actually replying to PJ_in_FL's post above mine. But the discussion has been interesting. It has given me cause to revise some of my non-Newtonian mechanics, and also brought back memories of that enormous field dismissed with the token word "ballistics" - still a very active R&D area in Defence science and technology. If only it were as simple as the schoolboy parabolic trajectory!

Alan

[ChrisGreaves]

I apologise for the volume. I got sidetracked by life this past week and have replied to all the pending posts in one horribly wrong epistle.


PJ
PJ> Consider the time and velocity of any object completing the circumnavigation of a point (e.g. center of The Earth). At any point in time that object has an X and a Y component of it's velocity.
I think there is a weakness here. I don’t know whether it invalidates the argument.
There is indeed an “X and Y component” of its velocity, but in what co-ordinate system?
For a low-velocity, short-duration flight (tossing a football in the air from my old hands) we can treat The Earth as a stable plane. But you will go on to treat The Earth as a rotating sphere, with the projectile’s initial reference at the time it leaves the barrel rapidly disappearing over the horizon.
I don’t want to get so far into maths that we use The Moon as an absolute point and calculate everything with “X and Y components” relative to the centre of The Moon.
PJ> ... the "X" portion (with respect to the center point) ...
Here too, if you are going to suggest cartesian vectors then I think you can’t have a “centre point”. You have to use polar coordinates (out of my depth soon) and radian measurements (now out of my depth)

Leif
Leif> ...it'll have to be done at either the North or South pole ...
Which I interpret to mean “we can safely ignore The Earth’s rotation”. A coward’s way out (grin).
Any solution should be accurate at the equator and at the pole and at all points in between, the poles being some sort of boundary condition.

AlanMiller
AlanMiller>... the ground rotates beneath the bullet ...
I’m getting out of my depth again. It is a fine point to argue that the ground rotates away from the bullet, rather than beneath it. “Away From” suggests a low-velocity low-altitude Flatlander scenario, whereas we are looking at what is probably a transition area between tossing a football and firing an ICBM.

PJ
... (their linear "X" velocities are the same) ...
My point above on Post 213751. I think that you are using two different frames of reference again, one cartesian and one polar.
It’s late at night, though, and my brain hungers for Bill Bryson and Garrison Keillor ...
I do trust all your calculations, if only because I didn’t specify any initial quantities.
I am having trouble visualizing The Earth rotating and the bullet rising, and their positions relative to each other.
I note that NASA doesn’t launch a rocket straight up away from The Earth. Well, only for the first fifteen seconds or so, then they roll it over onto its back and send it on a trans-atlantic tour. That is, rockets spiral up unto orbit.
Maybe we DO have to be Rocket Scientists!
PJ> Were I Issac Asimov writing one of his wonderful science essays that I ate up as a young lad, I might consider the effort. But for Chris, well ...
You CAN compare me to Asimov. He is like any one of his books. I am the mosquito squashed between page 71 and 72! (which is, of course, impossible ...)

AlanMiller
Alan> That sounds like a Coriolis-type correction to me.
I know that winds are above the plane of The Earth, but if I think of them slipping across the surface of The Earth, then from what I remember of Mr deKurloi’s geography lessons, the Coriolis effect on winds occurs when a wind moves with a component of its velocity towards or away from the poles.
That is, a body of air flowing from (roughly) new York to Florida will appear to veer westwards(?) relative to a straight line drawn from New York to Florida.
And you all know what I mean by a “straight line”, I mean a piece of string stretched taut between the two cities, curved to fit the shape of The Earth.

PJ
I wish your image didn’t look like winds creeping across the surface of The Earth instead of a bullet rising above The Earth. I’m trying to keep an open mind here.
PJ> ... is being applied in different reference frames so the end point location of the gun and the bullet will be different after a period of time.
This is so seductive. “different reference frames” appeals to me, but I still have a nagging feeling that the bullet is held by gravity and so is in an orbit-like trajectory. I know it is a parabola relative to The Earth’s surface, but it seems orbital with respect to a fixed point on The Earth’s surface.
(I must remember to sign this post “Confused” of Toronto)
PJ> The ball bearing will land eastward of the center of the base
I think I agree with this. You were saying in Post=213806 that the rifle, travelling upwards, falls behind the barrel, but the ball bearing, traveling downwards, falls ahead of the tower. That makes relative sense.

AlanMiller
Alan> ... so small that we can consider its surface to be a non-rotating, moving plane ...
Which is the case when I toss a football into the air. We can consider The Earth as an immobile plane. However this argument can not, surely, apply to an ICBM fired straight upwards (except for angling a bit for atmospheric variables). Admittedly you’d have to be an idiot to fire an ICBM straight up and hope that it would come back and visit your cafeteria, still ...
I can imagine (no atmosphere) firing a projectile upwards, at right angles to the sphere so that the sphere had time to rotate through ninety degrees before the projectile returned to the surface. At that scale my mind shrugs and supposes it can accept lateral differences.
Then I think about the LLM rising up to rejoin Apollo. Did that just shoot vertically off the surface of The Moon and trust/hope/calculate that the orbiting Apollo would appear over the horizon at just the right time and distance to effect a meeting? If so, shouldn’t we draft PJ to argue the travel in the opposite direction? I mean, the LLM is like the bullet going “straight up”.

PJ
PJ> So with that mind problem set up,...
I rather like this experiment. We ignore The Moon’s path around The Earth, and just consider The Moon as a rotating sphere.
This almost makes sense to me. I still have trouble visualising what the projectiles so-called horizontal component of velocity is wrt The Moon.

AlanMiller
Alan> ... it would mean a refresher course of my Lagrangian mechanics ...
And there, Gentlemen, I think I will leave you!

Thanks to all who contributed, especially PJ and Alan. I learned more than I had anticipated.
I’m also glad that we focussed on this Radial/Cartestian/Coriolis effect and not any other factor that I had failed to think of.

(signed) “Confused” of Toronto