Post by: Hayseed on 17/10/2019 21:49:47
If we had one hundreds weights spinning and cut them, Would they spread out in a plane or would they spread out in a cone?
Post by: Janus on 17/10/2019 22:37:25
As we tether out weights to slow the spin of a spacecraft, and then release the weights, does the path of the weights have the spacecraft's axial velocity component? Or only the radial component?A plane relative to the ship and a cone relative to someone with an axial motion relative to the ship (It doesn't matter if we consider this person or the ship as being the one moving.)
If we had one hundreds weights spinning and cut them, Would they spread out in a plane or would they spread out in a cone?
Post by: Hayseed on 17/10/2019 23:15:16
Post by: alancalverd on 18/10/2019 00:09:52
Post by: Hayseed on 18/10/2019 15:33:00
That is a ratio of 22,000 to 25. That's quite a ratio. As we tether out, the RPM will drop, making to ratio even higher.
What would the cone angle be? It would be super sharp, wouldn't it? Does this sharp cone concept actually occur? It should be noticeable.
Would it take a ratio of 1 to 1 give a cone angle of 45 degrees from craft, 90 degrees total?
What ratio would be needed for a 50/50 direction?
Has this been confirmed? This doesn't seem right. Does the craft velocity change upon release?
I would love to see this dynamic.
Let's say the final spin RPM is 5 RPM, with a 100 ft tether. That's about 53 ft/sec. It did increase a little.
22000/53.
Post by: alancalverd on 18/10/2019 17:25:12
Post by: Hayseed on 18/10/2019 18:14:37
Shouldn't there be a difference?
If we replace gravity for the rotational diameter tension, The bomb's cross velocity starts at zero.
With rotation, the cross velocity has already been established.
Should not this change the dynamic?
Post by: alancalverd on 18/10/2019 19:52:53
Post by: Hayseed on 18/10/2019 20:29:17
Consider this. Let's say that one has a special pistol and a special rifle. A .22 LR in the pistol, produces the same velocity as a .22 short in the rifle.
The only difference between them, is the acceleration duration to reach velocity.
We fire both from moving trolley, same time,same targets, will they have the same trajectories?
It does not matter if the bullets leave barrel at same time, it does not matter if bullets hit same spot.
What matters is, can one lay one path on top of the other? If they vary, does not this imply, that the trolley velocity, was added during the acceleration?
Post by: alancalverd on 18/10/2019 23:28:20
It is oddly counterintuitive that if you fire a bullet horizontally and drop a bullet from the same height at the same time, they will hit the ground at the same time. This is particularly important for long-range sniping where the bullet may fly for around 2 seconds, during which it will fall 64 feet. "Aim high" is more than just encouragement.
Post by: Janus on 18/10/2019 23:38:55
The only reason it curves, is because of the gradual effect of gravity. There is no gradual application of velocity with a spin.If you do this experiment with a trolley at rest, the bullets obviously follow the same trajectory if their velocities are the same.
Consider this. Let's say that one has a special pistol and a special rifle. A .22 LR in the pistol, produces the same velocity as a .22 short in the rifle.
The only difference between them, is the acceleration duration to reach velocity.
We fire both from moving trolley, same time,same targets, will they have the same trajectories?
It does not matter if the bullets leave barrel at same time, it does not matter if bullets hit same spot.
What matters is, can one lay one path on top of the other? If they vary, does not this imply, that the trolley velocity, was added during the acceleration?
Having the trolley moving doesn't change this. It is just as valid to consider the trolley at rest and you are moving. If the bullets follow the same trajectory according to the trolley, they can't follow different trajectories as viewed by someone with a relative velocity with respect to the trolley
Post by: Hayseed on 20/10/2019 05:51:34
No gravity in discussion. Gravity will not allow a constant velocity.
There is ONE acceleration in this setup(bullet), and it is of short duration. And you did not catch my question about the duration of it. So forget that aspect.
If we trade out the gun set-up for the rotation set-up..........then there is no velocity acceleration at all in the setup. If the acceleration is removed, will it change the trajectory, and will it change the impact velocity?
Can you follow along with what I am asking?
Post by: Hayseed on 20/10/2019 14:54:23
At a 100 ft radius, the perimeter would be 628.31 ft. The final RPM is 5 RPM. That's 3141.59 ft. per min. Divide by 60 sec.......52.35 ft/sec.....cut cross velocity.
Are these figures correct?
Let's say I have a gun with 1000 ft/sec. Let's say I have a ball in rotation at 1000 ft/sec.
I cut and fire gun........which one hits target first?
Post by: Janus on 20/10/2019 16:02:19
A diameter of 3 ft matches a perimeter of 9.42 ft. !50 RPM on 9.42 ft is 1413.7ft. per min.........and that divided by 60 is 23.56 ft. per sec. Is that correct?As long as the the gun is pointed along the tangent of the rotation so that at the moment it is fired and the moment the ball is released, both ball and gun projectile are headed in the same direction, they will travel side by side and hit the target at the same time.
At a 100 ft radius, the perimeter would be 628.31 ft. The final RPM is 5 RPM. That's 3141.59 ft. per min. Divide by 60 sec.......52.35 ft/sec.....cut cross velocity.
Are these figures correct?
Let's say I have a gun with 1000 ft/sec. Let's say I have a ball in rotation at 1000 ft/sec.
I cut and fire gun........which one hits target first?
So if the ball is moving along the red circle at 1000 ft.sec, and is released at the top, it will follow the red arrow at 1000ft/sec. As long as the gun is pointed in the same direction ( to the right here), is anywhere on the green line, and is fired at 1000 ft/sec, at the same time as the ball is released, it will follow the blue line and remain constantly abreast of the ball.
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Post by: Janus on 20/10/2019 17:30:43
A diameter of 3 ft matches a perimeter of 9.42 ft. !50 RPM on 9.42 ft is 1413.7ft. per min.........and that divided by 60 is 23.56 ft. per sec. Is that correct?A suggestion:
At a 100 ft radius, the perimeter would be 628.31 ft. The final RPM is 5 RPM. That's 3141.59 ft. per min. Divide by 60 sec.......52.35 ft/sec.....cut cross velocity.
Are these figures correct?
Let's say I have a gun with 1000 ft/sec. Let's say I have a ball in rotation at 1000 ft/sec.
I cut and fire gun........which one hits target first?
As Halc noted in his post, he was assuming rotations/sec, not RPM. Generally, in physics, the second in used as the base time unit. So there is a tendency to automatically think in these terms. Rotations per sec (or even radian/sec) are more "Natural" to work with.(Radians per sec has the advantage of being directly converted to tangential speed. n radians per sec with a radius of x meters gives a tangential speed of n*x meters/s)
You should also learn to start using the MKS( Meters-kilograms-sec) system of units rather then units like feet. Physical constants tend to be given in these units, and it is easier to work directly in them rather then having to convert between unit systems.
Post by: Hayseed on 21/10/2019 15:16:40
That diagram shows when the cut release is at the same time as muzzle exit, not firing time.
Let's try something different. Imagine an American football field. On the close side of the field we have a trolley, and on the other side a wall with the field yard lines extended. The field lays east-west. Trolley at south, wall at north. With trolley stationary at 50 yard line, we shoot wall along the line....hit wall. No gravity, no air. The bullet will strike wall 50 yard line at muzzle velocity.
Are we all together?
Now, the trolley is moving from right to left, and we fire incident at 50 yard line point. OOPS....I shouldn't say fire.....the bullet leaves muzzle at 50 line incidence.
Does the bullet hit the line?....or veer to the left?....or veer to right?
If we keep the muzzle v constant, and vary the trolley v, will the bullet always land on the same side of the 50 wall line. Even if trolley v, exceeds bullet v?
Post by: Janus on 21/10/2019 17:02:04
Ok, I will try to use metric. Evidently, no one is concerned with period of acceleration.
That diagram shows when the cut release is at the same time as muzzle exit, not firing time.
Let's try something different. Imagine an American football field. On the close side of the field we have a trolley, and on the other side a wall with the field yard lines extended. The field lays east-west. Trolley at south, wall at north. With trolley stationary at 50 yard line, we shoot wall along the line....hit wall. No gravity, no air. The bullet will strike wall 50 yard line at muzzle velocity.
Are we all together?
Now, the trolley is moving from right to left, and we fire incident at 50 yard line point. OOPS....I shouldn't say fire.....the bullet leaves muzzle at 50 line incidence.
Does the bullet hit the line?....or veer to the left?....or veer to right?
If we keep the muzzle v constant, and vary the trolley v, will the bullet always land on the same side of the 50 wall line. Even if trolley v, exceeds bullet v?
bullet.png (25.1 kB . 480x418 - viewed 3741 times)If the bullet leaves the gun perpendicular to the path of the trolley ( as measured from the trolley), it will follow the path indicated in the diagram, and strike the wall/target at a point exactly opposite of where the trolley is when the bullet hits ( as shown by the cyan line). The relative speed difference between trolley and bullet makes no difference.
Post by: Hayseed on 21/10/2019 17:15:55
Post by: Halc on 21/10/2019 17:57:34
ok, what happened to the bullet's velocity?Bullet velocity is the vector sum of trolley velocity and muzzle velocity relative to the trolley. This is true in any frame.
Janus must be an Aussie because he drew the North wall at the bottom. :)
Post by: Janus on 21/10/2019 18:17:39
Actually, I didn't pay any attention to the cardinal directions (they really don't matter in the final analysis). I originally started drawing the diagram with the trolley going from bottom to top, changed my mind, and rotated it. I had already added some of the labels, and quite frankly, was too lazy to start over.ok, what happened to the bullet's velocity?Bullet velocity is the vector sum of trolley velocity and muzzle velocity relative to the trolley. This is true in any frame.
Janus must be an Aussie because he drew the North wall at the bottom. :)
Post by: Hayseed on 21/10/2019 18:26:52
Let me ask you this. And thanks to all for participating.
Does the flight time change?
Post by: Janus on 21/10/2019 18:46:49
ok, what happened to the bullet's velocity?Bullet velocity( along black arrow) would be equal to sqrt((trolley velocity)^2+(muzzle velocity)^2)*
* to a good approximation when trolley velocity and bullet velocity are low compared to c.
If they were significant fractions of c, you'd need to use
sqrt(u^2+v^2- (uv)^2)
where u is the speed of the bullet as measured from the trolley, and v is the velocity of the Trolley, expressed in units of c.
(Actually, this is the correct equation for all cases, it is just that for low values of v and u the answer comes out to being almost exactly what you get using the first equation above)
The bullet still remains abreast of the trolley during its entire flight.
Post by: Janus on 21/10/2019 18:55:05
Ok, so my target moves parallel to the trolley, the further I move away from trolley(staying parallel), the faster the bullet travels?The distance between target and trolley path has no effect on the bullet speed. It still travels at the same speed, it just travels a longer distance and take more time to hit the target. It is no different than if the trolley and target were not moving with respect to each other. if the bullet leaves the gun at 100 m/s and the target is 100 m away, it takes 1 sec for the bullet to hit the target, if the target is 200 m away, it takes 2 sec. Having either the target or trolley moving at some velocity perpendicular to the direction the gun is aimed has no effect on this time.
Let me ask you this. And thanks to all for participating.
Does the flight time change?
Post by: Hayseed on 21/10/2019 19:59:25
Post by: Halc on 21/10/2019 20:21:07
I start a clock when the bullet clears muzzle. I stop that clock when bullet hits wall. Is there ever a time difference?No, not as long as the muzzle is some fixed distance from the wall. In post 22 you use wording that suggests the shooting goes on in a place that is moving away from the trolley, and is thus somewhere else.
So in the diagram in post 18, the speed of the trolley has no effect on the bullet travel time.
Post by: Hayseed on 21/10/2019 20:52:04
Are you sure that the flight times don't change?
Does the trolley v........impart speed on bullet, or direction on bullet?
Post by: Hayseed on 22/10/2019 00:14:39
After leaving the muzzle, we can not increase the speed of the bullet.
And because of the cross displacement, a longer flight path and time.
If we use the magnitude trolley v for side b, and the magnitude of bullet v for the hypotenuse.......the resulting side a........will be the resulting magnitude of velocity of incident impact.
The side a, will always be shorter than the hypotenuse(muzzle v)........less velocity.
Does the impact hole in wall, go straight in? Does that hole go deeper with trolley velocity? The bullet's mass remains constant.
How does this really work?
Post by: Janus on 22/10/2019 01:41:30
Most take one v as side a and the other v as side b, with the hypotenuse as resultant v.If the gun is fired while pointing in the A direction, then the A side is the muzzle velocity and the the resultant velocity( along the hypotenuse) is the vector addition of the trolley velocity and muzzle velocity. It will be greatwer than either trolley or muzzle velocity. If muzzle velocity is 100 m/sec and trolley velocity is 15 m/sec, then the velocity along the hypotenuse will be ~101.119 m/sec. The bullet will strike the target at this speed and at an angle of ~8.627 degrees from straight in.
After leaving the muzzle, we can not increase the speed of the bullet.
And because of the cross displacement, a longer flight path and time.
If we use the magnitude trolley v for side b, and the magnitude of bullet v for the hypotenuse.......the resulting side a........will be the resulting magnitude of velocity of incident impact.
The side a, will always be shorter than the hypotenuse(muzzle v)........less velocity.
Does the impact hole in wall, go straight in? Does that hole go deeper with trolley velocity? The bullet's mass remains constant.
How does this really work?
What makes you think that the hypotenuse velocity would be equal to the muzzle velocity?
Post by: Hayseed on 22/10/2019 06:35:35
Is that correct?
How would one calculate the bullet v, if we aim gun at 45 degrees?