Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: Hayseed on 17/10/2019 21:49:47

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?
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?

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?
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?
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.)

Will the ratio of the axial velocity to the rotational velocity vary the angle of the cone?

Yes.

Alrighty, what is the velocity of the craft? let's say 15,000 MPH(22,000FT/sec). Lets say the diameter of craft is 3 ft. And rotating 150 RPM(2.5 rps). That's about 25 ft. per second on surface of craft.
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.

No need to go into space to do this. When you cut the string, the weight will fly off in a straight line at 25 ft/sec away from the axis of the spacecraft and 22000 ft/sec in the forward direction. This is standard ballistics for moving vehicles if you ignore air resistance and gravity (which you can in space, of course!) . Gunners and bomb aimers spend a lot of time drawing such triangles in their studies.

So, are you saying that dropping a bomb from a moving plane, has the same dynamic and trajectory effect as the cut of rotating weights in space?
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?

No big deal except the bomb accelerates downwards until it reaches terminal speed, so one side of the "triangle" is curved. Glide bombs and parachute mines have different trajectories but the principle is always the same  vector addition.

The only reason it curves, is because of the gradual effect of gravity. There is no gradual application of velocity with a spin.
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?

lf the muzzle velocities are the same, and the muzzles are level with each other, the trajectories will be the same.
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 longrange sniping where the bullet may fly for around 2 seconds, during which it will fall 64 feet. "Aim high" is more than just encouragement.

The only reason it curves, is because of the gradual effect of gravity. There is no gradual application of velocity with a spin.
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?
If you do this experiment with a trolley at rest, the bullets obviously follow the same trajectory if their velocities are the same.
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

ok, there is a lot of diversion here, from what I am talking about. The reason I used a spacecraft and a trolley, was to exclude gravity. Gravity causes a velocity, but it is an acceleration. A trolley has constant velocity. A constant velocity can not be had with gravity. No need to divert with a comment about terminal velocity....save it for tea.
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 setup for the rotation setup..........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?

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 setup for the rotation setup..........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?
Your question is not hard and simply requires vector addition.
You ask about impact velocity of a bullet, but you've described neither what is impacting (or in particular, its velocity relative to the ship), nor the speed of the bullet as it leaves its gun, so that cannot be answered.
Velocity is frame dependent, so the answer to your original question (the shape of the path of the weights) is also frame dependent.
Alrighty, what is the velocity of the craft? let's say 15,000 MPH(22,000FT/sec). Lets say the diameter of craft is 3 ft. And rotating 150 RPM(2.5 rps). That's about 25 ft. per second on surface of craft.
You've chosen a frame where the ship is moving at that speed and presumably in the direction of its axis of rotation. Then yes, while the computation of the speed of the surface of the craft is wrong (it's about 9.5 ft/sec), a series of objects released from that surface would spread out in a narrow pointy cone. In the frame of the ship (regardless of orientation of spin), the objects will spread out in a flat plane. In that frame, the ship has zero velocity.

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?
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?

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?
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 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.
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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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?
Sorry, yes. I seemed to be using 1 rotation per sec.
At a 100 ft radius, ... 52.35 ft/sec.....cut cross velocity.
Are these figures correct?
Sure.

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?
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?
A suggestion:
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( Meterskilogramssec) 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.

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 eastwest. 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?

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 eastwest. 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 316 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.

ok, what happened to the bullet's velocity?

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. :)

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. :)
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, so my target moves parallel to the trolley, the further I move away from trolley(staying parallel), the faster the bullet travels?
Let me ask you this. And thanks to all for participating.
Does the flight time change?

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.

Ok, so my target moves parallel to the trolley, the further I move away from trolley(staying parallel), the faster the bullet travels?
Let me ask you this. And thanks to all for participating.
Does the flight time change?
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.

Ok, so my target moves parallel to the trolley, the further I move away from trolley(staying parallel), the faster the bullet travels?
Let me ask you this. And thanks to all for participating.
Does the flight time change?
The flight time does not change due to the motion of the trolley (which moves parallel to the wall and thus has no effect on its distance from it). But you're talking about you moving yourself away from the trolley? That makes no difference if you're not involved in the gun being shot from the trolley.
If you're firing the gun and the gun is distancing itself from the target, then of course the time to get there is going to go up, which is what Janus answered. What the trolley does in that case is of no concern since it isn't involved in the gun (not on the trolley) firing at the wall.

I start a clock when the bullet clears muzzle. I stop that clock when bullet hits wall. Is there ever a time difference?

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.

Sorry about that, I meant the target moving away.
Are you sure that the flight times don't change?
Does the trolley v........impart speed on bullet, or direction on bullet?

Most take one v as side a and the other v as side b, with the hypotenuse as resultant v.
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?

Sorry about that, I meant the target moving away.
If the trolley is moving parallel to the wall, then the wall (target) does not get further away, and the time is unaffected by trolley speed. So I'm unclear what you mean by the target moving away. If the trolley is moving (parallel) relative to the wall, then the bullet hits directly across from where the trolley is at impact time (see pic in post 18), and if you consider that impact point to be the target, then the impact point actually is closest to the trolley at that moment, so it can be argued that the target is approaching, not moving away.
Are you sure that the flight times don't change?
Not if the motions involved are parallel like you describe.
Does the trolley v........impart speed on bullet, or direction on bullet?
Both. Again, see pic in post 18. The direction is diagonal given trolley motion, and the speed needs to be greater to get there in the same time as the stopped case. The velocity of the bullet is the vector sum of the trolley velocity and the muzzle velocity, so if the trolley is moving, that sum will be greater than the case where the trolley is stationary and adds nothing to the vector sum of the bullet.
Most take one v as side a and the other v as side b, with the hypotenuse as resultant v.
That's how vector addition works yes, at least at low speeds. We're not talking relativistic speeds here, right?
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.
a is the muzzle velocity of the gun (relative to the gun). Hypotenuse is the bullet velocity relative to the frame you've chosen, the one in which the trolley has velocity b.
The side a, will always be shorter than the hypotenuse(muzzle v)........less velocity.
Not for all triangles, but for the right triangle you're describing, yes, this is true. If I fire the bullet somewhat behind the train, the bullet might be going slower than the muzzle velocity since the vector sum of the two will be less magnitude than velocity a by itself.
Does the impact hole in wall, go straight in? Does that hole go deeper with trolley velocity? The bullet's mass remains constant.
Not straight in if the trolley and wall are not stationary relative to each other. You see the angle in the picture in 18. It goes in at that angle, and presumably makes a deeper hole due to it going faster

Most take one v as side a and the other v as side b, with the hypotenuse as resultant v.
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?
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.
What makes you think that the hypotenuse velocity would be equal to the muzzle velocity?

Ok, I am following what you are saying. So if the trolley v is equal to the muzzle v, then the departure angle would be 45 degrees from the trolley.
Is that correct?
How would one calculate the bullet v, if we aim gun at 45 degrees?