OK I just saw part of it where they mention “spin drift” which may be the Coriolis affect.
Yes, there would be an effect. It may help to take a globe and pick a spot at, say, 45 latitude, and try to visualize where a shot fired due east ends up as the globe rotates through a greatly exaggerated amount of distance (since it will be hard to visualize if you try to turn a globe for an equivalent angular rotation which would occur over 2 seconds).
If they’re talking about gyroscopic drift, which is often called spin drift, that’s a different phenomenon. I don’t know a good shorthand way to calculate spin drift, and Cecil didn’t ask, so I let that sleeping dog continue dreaming happily away.
No, I don’t remember hearing anything about that.
The factory Win .308 is considered marginal at 1000 yards. The Marines load them hotter or use more specialized rounds like the .50 BMG out of a Barrett or the .338 Lapua. There is also a new .416(?) that has amazing external ballistics.
re: effect of heat on a hollow, round object:
There was an old riddle about this at Games magazine: if a ring with finger size X (the hole inside the ring) was heated to a point where the metal expanded Y%, what would be the effect on X? (Would the hole get smaller or larger?)
At the time, the author’s answer (supposedly a physicist, although many readers disagreed), was that the interior of the hole would remain the same size. It was hotly debated in later letters columns, but I don’t know what the real answer is today. My memory says that the metal expands in all directions equally, therefore, as the interior of the circle expands inwards, the expansion of the outside of the ring pulls it out proportionately, thus the hole remains the same size. Iirc, the author noted that married people don’t get their fingers pinched off when they step out of the shower, for example.
re: Coriolis effect on driving
Ah, is this why you have to move the steering wheel slightly even though the car has perfect alignment?
I concur. The reason for picking 1,000 yards was partially to exaggerate the Coriolis effects.
I can’t account for all situations, but very often that is due to the crowning shape of the road, which serves to drain water off to the sides.
I suppose that makes sense, considering you used Remington’s calculator, that it would use Remington’s own Winchester .308.
Looking at your bote calculations, it appears as though you’re using muzzle velocity to calculate drop at 100 yds and at 1000 yds. For the Remington .308 Winchester (150gr) at 100 yds, it would be closer to 2500fps and at 1000yds, 500fps. You should be using average velocity for the distance, no? For your 1000yd example, that’s about a 3 inch difference.
You wouldn’t expect a ring that was fitted cold to pinch your finger off just because it’s returned to its nominal temperature after being warmed slightly, now would you? A shower heats metal by only a few tens of degrees. And against that observation there are many practical observation concerning, say, how blacksmiths fitted iron tyres to cartwheels (heat 'em red hot, pop them on, contract to fit).
Consider a long thin block expanding when heated. It lengthens by the same proportion in all directions - which means that the greatest increase is in the direction of the greatest dimension, the length, and the increase in thickness is hardly any. Now bend that block into a ring and start again. Clearly by far the greatest increase is in the circumference of the ring, and hence the diameter, and this will swamp the decrease in diameter caused by the increase in thickness. And I’m not a physicist, just a poorly-qualified applied mathematician.
I went back and worked through the calculations (from here and here), simplifying for the extreme case where a layer of metal on the inside is suddenly much hotter, and the diameter of the outside of the barrel is comparitively large. In that case, a lot of the equation simplifies out, and the upshot is that with a temperature increase where the unconstrained increase of the barrel ID (d[sub]i[/sub]) is [symbol]aD[/symbol]Td[sub]i[/sub], the constrained increase of the barrel ID is [symbol]aD[/symbol]T(1-[symbol]n[/symbol])d[sub]i[/sub], where [symbol]n[/symbol] is Poisson’s ratio which is always less than one, and is specifically 0.3 for steel.
In other words, even under uneven heating, the barrel ID will always grow.
It’s possible the barrel will ovalize or some such under uneven heating, but the original statement that “…as a rifle is fired, its barrel heats up, the metal contracts and the bullets are propelled faster” is bogus (assuming I did the math right).
You may be misunderstanding - I’m using the flight time to target, as calculated by Remington. This took into account the altitude and a “standard” temperature for Sacramento which I assumed. The muzzle velocity actually does not appear in my calculations. I did calculate a mean velocity, which is simply distance over the Remington calculated flight time.
Yes but at a 1000 yards? I admit I’m crap at physics and I don’t know the curvature of the earth so I’m just assuming, but I’d imagine that at that distance it would be negligible and that local geography (hills and ditches etc.) would be far more significant differences in latitude.
It was well known to artillerymen in the muzzleloading, black powder era that, as a cannon was repeatedly fired, it would heat up and the recoil and range would both increase. What the scientific cause behind this is, and whether there is a similar effect when firing a modern rifle, I leave as an exercise to the reader.
Also, to what effect is the differing rotational speed of firer and target offset by the fact that the air through which the bullet is travelling is also moving sideways?
I believe you do:
1 – 2*(rotation of the earth, rad/s * muzzle velocity / G)*cos(latitude)*sin(azimuth)), where azimuth is measured clockwise from due north.
Ditto for putting tires on locomotive driver centers, and those are inches thick, not a little, thin wagon wheel rim. The sight is quite spectacular.
My high school physics class had a brass ring on a wooden-handled stick, and two brass balls, ditto. At room temperature one ball would pass the ring but not when it was heated. The other would not pass the ring at room temperature, but would if the ring was heated. It only took about thirty seconds in a bunsen burner flame to make the difference.
Here’s a mental image that works for me: take a picture of the ring and put it in a copier and blow it up by 100%. Do it again a few times until the picture of the ring is 2 feed wide. The interior diameter of the ring (in the new picture) will be much larger than the original picture.
Oh sorry, I thought you were talking about the horizontal effect. There was a reason I had to use the muzzle velocity for the vertical equation, but I’m not sure I recall what it was, although I admit freely on first glance it is confusing. I’ll have to check my notes and see what the reason was.
Every point along the same line of longitude will have the same angular velocity e.g. 1 rotation per 24 hours. The only thing that will differ is linear speed because a point along the equator has to move farther then a point closer to the poles in the same time period. But you don’t see continents shearing off of themselves near the equator because they are trying to “rotate faster” then their more poleward sections do you? This is where relativity comes in. As far as a directly eastward shot say, the bullet will travel at its velocity x + earth’s rotational speed y, but the target will be traveling at this speed y as well so only the bullet’s speed has any effect on the path of its travel.
In the barrel heating up unevenly you have to distinguish between a single shot and multiple shots. In a single shot the bullet leaves the gun long before the heat generated by it has a chance to significantly affect the barrel. In multiple shots, the barrel would be uniformally heated after a while so the only time it would make a difference is if you fired it in bursts, separated by long periods of not firing it (so it could cool back down)
Going back to the column, surely the car won’t be deflected, since it’s not moving independently of the rotating reference frame.