Oh, it’s been proven. Long ago. Obviously, not to your satisfaction, but it’s well established. Besides, if 0.999~ doesn’t equal one, how do you express the difference between the two?

No, it’s because it’s true. Ask any mathematician. If you divide 1 by 3, you get an infinite series of repeating threes after the decimal point. Which is exactly what 0.333~ represents.

Hope you don’t like your house too much; you’re about to get schooled.
Powers &8^]

Ok, so show me the proof! Not a convention, not a convenient tool we can use so our brains don’t explode. What’s the indisputable truth? And what about the calculus? Come on guys it’s just a bit of fun but I’m inviting comments here. If you think we need a new thread then so be it…

No, I am specifically eliminating factors that are irrelevant because we compensate for them in other ways or they don’t change. The model I describe is exactly how I visualized it myself to correct my own errors as discussed before.

The problem is that you are not quite following my description closely. I am not firing in a random direction. However, it is possible my description does not adquately explain the analogy.

The rotating chair works as the rotating earth, with the axis of the chair serving as the N-S axis of Earth. The chair rotates horizontally rather than the vertical rotation one would experience on the equator. I mounted the target to the far chair arm just like the target would be standing on the Earth’s surface and thus constraint. Additional movement could be included in the math, but it would be unnecessarily more complex and mask the effect in question.

The reason why I picked an arbitrary distant point of reference was so I could define the line between my eyes and the target at time zero. That line is the path the “bullet” takes. It defines the spot on the target at the time of firing. My problem was trying to visualize the Earth rotating and being on the surface and seeing which direction the actual change occurred. That is why I had to pick a distant point to be the reference line.

The key to the problem is that the bullet takes time to travel, and during that time the target changes position. The key factor, once again, is that this is occurring on a curved surface, so the rotation does not cause one vector, it causes two. The speed of rotation is defined by the angular velocity and the distance from point of rotation. That speed remains constant throughout the bullet’s flight, and is the component that you are visualizing. That element is zero effect because the target, the shooter, and the bullet are all affected the same way by that angular rotation.

However, the component you are missing is caused by the angular position change. If the surface were a flat plane, then moving forward or backward in the plane would be the only affect. But because the surface is curved, the effect is that there is a hump between the shooter and the initial target location. The path that the target moves is not in a straight line, but a curve along that hump.

When the bullet reaches the distance to the target at time 1 (the point of impact), the target has moved forward some short distance, and by doing so, has moved up the hump some short distance. That movement up the hump makes the target spot (say center of ring) actually a couple inches higher than where the bullet comes along.

zut said:

Thank you! I was trying to work up an example.

From one end of the train to the other as it comes around the curve, there is a horizontal displacement of the end of the train from your perspective at the front. That horizontal displacement is the same as what is occurring with the curvature of the Earth’s surface and the rotation.

It’s like skeet shooting - if you fire the skeet directly away from the shooter, it is relatively easy to hit because you just fire downrange and the shot will eventually hit the clay pigeon. But if you angle the vector of the clay pigeon, then the shooter must account for the horizontal component of the flight, and has to “lead” the flight. You fire, the pigeon moves as the shot races after it. If you don’t account for the horizontal vector, then the spread pattern on the shot may not be enough to compensate and you will miss the shot. Only by aiming ahead of the pigeon can you put the shot where the target will be.

Why the fuck would you bring that up in this thread? We can’t get people to agree on that in the math threads, even when we show them mathematicians saying it. If you’re trying to explain a contentious point, for Og’s sake don’t use another contentious example!

Homer7 said:

Shot is from West to East, i.e. into the Earth’s rotation. The target is moving up the curve of the path of rotation. The target spot moves up, the bullet will hit low.

Negative, you just argued that speed of the target, shooter, and bullet are all being affected the same by the rotation. The Earth’s surface and the air are all also affected the same by that speed. Ergo, it has no effect on the flight time or bullet’s path. There is no change to the effect of the air.

Not at all. 3 will make the bullet hit low. 4 will have no change (whatever effect the air has will be the same). 1 will tend to make the bullet hit high (because of parabolic flight). There is a tiny drag component from sliding through slightly less air because the total distance between the shooter and the target is slightly covered by moving the shooter.

T= target
S= shooter
The distance traveled by the bullet is the 15 dashes between S0 and T1.

NOTE: this doesn’t account for curvature and is only used to display the distance effect of the motion. I’m not talented enough with ASCII art to attempt a curved path.

Offhand I can’t say which has a greater effect - motion by curvature, parabolic path, or decreased drag. My gut says drag is least significant, but I have no real basis for that call.

I can’t help you (especially Mr Zut) so I give up, this will be my last post. What a fine cloak of anonymity, the internet, for those wishing to show their true colours. ‘Lets gang up on the newbie - he’s questioning our beliefs!’ Well yes actually I AM questioning them AND your methods. Let’s have a little more proof, or reference to published material, or to something tangible. Thought experiments are very useful but they can be deeply flawed.

There’s an awful lot to comment on here but I can’t be bothered. I will continue with one point though. Quickly:-

I admit I was wrong to over simplify in my first post. But my reasoning is this: NO SNIPER ever had to worry about the motion of the Earth, it’s effect is too small. However, EVERY AIRLINE PILOT knows all about it! Let’s find an example:

If I fly from London to Istanbul, W-E and hence along with the motion of the Earth, it takes a few minutes longer than the return trip as the target (Istanbul) is receding. If the pilot could magically stop the plane and hover at a fixed point in space he would see that not only is Istanbul moving away from him but also DROPPING DOWN, disappearing below the horizon if left long enough. It’s exactly the same for the bullet. Here’s my proof - look at your airline tickets or look up flight times for any flights with a significant W-E component and you will see.

Oh, and:

0.3 recurring is, for all practical purposes, equivalent to 1/3
but 0.3 recurring is not 1/3, they are different abstract concepts.

0.9 recurring is, for all practical purposes, equivalent to 1
but it is not 1.

If you’re interested there’s a wonderful book - ‘Infinity’ by Brian Clegg. ISBN 978-1-84119-650-3.

Although I appreciate you might feel “ganged up on,” a) there’s no particular malice behind pointing out that you’re factually wrong, and b) if nothing else, Irishman spent quite some time and effort patiently explaining the physics to you, and you’re kind of blowing him off.

I think it’s only fair to point out that I’ve personally included at least one reference in this very thread, and Homer7, who’s calling for “a little more proof,” has provided exactly zero references. Why that might be is left as an exercise for the reader.

You are correct that to a stationary observer, Istanbul is moving away and dropping down. This is exactly the effect that I am talking about, why the target is moving out of the path of the bullet.

You are incorrect that the bullet will fall along with the target. If this were the case, then Coriolis acceleration would not exist. But Coriolis is real, affects airplanes and hurricanes, and even long range gunfire - definitely ship to ship weapons or artillery.

The bullet is affected by the airflow, which is sort of connected to the Earth. The bullet is affected by gravity, which causes the parabolic flight path. The movement of the target directly away from or toward the shooter is zero sum, no effect. But the rotation of the Earth means that the target’s final location is not the initial location, and it is not just changed by a linear position, but an angular position. That angular position change does not occur to the bullet in flight.

Since you requested a reference, I found one. I started with Coriolis on wikipedia and from there found this:

The effect has a name (Eötvös effect) and a formula to calculate the amount of change.

So, let me try some math:

2 v u + u[sup]2[/sup]
a = ____________
R

where

a = acceleration by Eötvös effect
v = velocity of Earth’s surface[sup]*1[/sup] = 465 m/s
u = velocity of bullet[sup]*2[/sup] = 2000 ft/sec = 609.6 m/s
R = Earths radius[sup]*3[/sup] = 6,378 km = 6,378,000 m

[sup]*1[/sup] Velocity of Earth’s surface taken from wiki cite above.

[sup]*3[/sup] Radius of Earth taken from wiki cite above.

Therefore,

2(465)(609.6) + (609.6)[sup]2[/sup]
a = _______________________
6,378,000
a = 0.147 m/s[sup]2[/sup]

What affect does that have on a sniper bullet? Assume a reasonable 500 m shot.

bullet flight time t = x/u = 500m / (609.6 m/s) = 0.8202 sec

distance caused by acceleration x-x[sub]0[/sub] = v[sub]0[/sub] t + 1/2 a t[sup]2[/sup]

d = x-x[sub]0[/sub]
v[sub]0[/sub] = 0 because we are looking at the independent effect of this one effect, which turns on as bullet leaves gun

ergo

d = 1/2 a t[sup]2[/sup]

d = 1/2 (0.147 m/s) (0.8202 sec)[sup]2[/sup] = 0.049 m = 4.9 cm
So, if my math is correct, for an average rifle bullet at a reasonable 500 m range shot, the effect of motion into or away from the direction of Earth’s rotation is a 4.9 cm drop or rise.