Down wind faster than the wind

[humber]

When viewed from the rear with a tailwind, the prop will always turn CW when the wheels are engaged with the rolling surface and moving downwind. This is without exception.

When the wheels are *not* engaged and moving downwind, (in other words, when you lift the cart up and hold it in a tailwind), the prop rotates CCW. This is without exception.

We and the cart both agree on the above. Also, the cart doesn't seem to care if someone other than us feels differently -- it still agrees with us.

JB

So when in full flight, the wheels are not on the ground?
ETA:
Assuming you agree that the prop is CCW in that case.

 

[humber]

humber, you seem to still have a problem with simple basic physics.

Lets take the person/vehicle collision and simplify it one step further by taking the ground away.

In case A, An astronaut floating in free space hits a space ship at 25 m/s.
In case B, The space station hits the astronaut at 25 m/s.

What is the difference between these two cases other than the frame of reference?

What difference does it make if there happens to be a surface we call the ground?

If a 1kg mass hits a stationary 10kg mass, the total momentum is much less than a 10kg mass hitting a stationary 1kg mass. (Same velocities)
Free space or not.
From the point of view of each of the masses, in each collision, their relative velocities remain equal from each viewpoint, but the actual events are different. They are not equivalent.

 

[Dan O.]

If a 1kg mass hits a stationary 10kg mass, the total momentum is much less than a 10kg mass hitting a stationary 1kg mass. (Same velocities)
Free space or not.
From the point of view of each of the masses, in each collision, their relative velocities remain equal from each viewpoint, but the actual events are different. They are not equivalent.

Here is the kicker. You are the personal injury attorney for the astronaut. Prove that it was the space ship that hit the astronaut and not the other way around.

 

[ThinAirDesigns]

So when in full flight, the wheels are not on the ground?
ETA:
Assuming you agree that the prop is CCW in that case.

I've never used the term "full flight" and I can only guess what you mean by it -- thus I can't agree with anything you said.

JB

 

[ThinAirDesigns]

Here is the kicker. You are the personal injury attorney for the astronaut. Prove that it was the space ship that hit the astronaut and not the other way around.

DanO, I do like how you think.

JB

 

[humber]

Here is the kicker. You are the personal injury attorney for the astronaut. Prove that it was the space ship that hit the astronaut and not the other way around.

Something tells me that their might be a problem equating the space-station, with the cosmic radiation that hits it, on the basis that they both have the same relative velocities.


1kg mass at 1m/s hits 10kg stationary mass. p = 1kgm/s

10kg mass at 1m hits 1kg stationary mass. p =10kgm/s

Not the same momentum. Different final velocities, too.

 

[humber]

I've never used the term "full flight" and I can only guess what you mean by it -- thus I can't agree with anything you said.

JB

During the period around 0:19 to 0:21, # 3. It looks like it is going CCW.
So, which way is the prop turning during this period?

 

[humber]

DanO, I do like how you think.

JB

QED

 

[ThinAirDesigns]

during the period around 0:19 to 0:21, # 3. It looks like it is going ccw.
So, which way is the prop turning during this period?

cw.

Jb

 

[Dan O.]

Something tells me that their might be a problem equating the space-station, with the cosmic radiation that hits it, on the basis that they both have the same relative velocities.


1kg mass at 1m/s hits 10kg stationary mass. p = 1kgm/s

10kg mass at 1m hits 1kg stationary mass. p =10kgm/s

Not the same momentum. Different final velocities, too.

To an observer traveling parallel to the space station, what are the final velocities for each case? Assume an elastic collision if you wish.

 

[humber]

cw.

Jb

Thanks. There certainly appears to be a period when it is running CCW, but aliasing makes that unsure.
It is possible that the wheels could simply spin. This would add drag, but if there is lift, it may not be significant. Without a comparable cart without gears to the wheel, it is difficult to say which would be the faster.

 

[humber]

To an observer traveling parallel to the space station, what are the final velocities for each case? Assume an elastic collision if you wish.

The collisions are different! They are different events, not different views of the same event. The result is completely independent of the viewer, all viewers. You can find different combinations that will yield the same final velocities. Do you require proof by exhaustion of calculation?
The final velocities will be what they are for each observer in each case.

The 10kg mass requires more energy to get it to 1/ms, than does a 1kg mass. What is there not to understand?

 

[ynot]

SPORK - Are you going to try the test I suggested in post 996 and let us know the results? You have all the equipment to do it and it would be very easy for you to do so.

 

[ThinAirDesigns]

SPORK - Are you going to try the test I suggested in post 996 and let us know the results? You have all the equipment to do it and it would be very easy for you to do so.

Ynot, I spoke with spork on the phone about that test this evening. We both agreed it's a fine test to do and will be done. I'll try to drop by his house before work in the morning and get it done.

I'm interested in what you think the difference might be in using the string to get up to speed, or my hand?

JB

 

[sol invictus]

The collisions are different! They are different events, not different views of the same event.

That "!" is poignant - it almost makes me pity you.

It's kind of touching to see someone struggling so desperately with reality.

 

[jsfisher]

The collisions are different! They are different events, not different views of the same event. The result is completely independent of the viewer, all viewers. You can find different combinations that will yield the same final velocities. Do you require proof by exhaustion of calculation?
The final velocities will be what they are for each observer in each case.

The 10kg mass requires more energy to get it to 1/ms, than does a 1kg mass. What is there not to understand?

So how do you establish which one is moving, astronaut or space ship?

 

[humber]

That "!" is poignant - it almost makes me pity you.

It's kind of touching to see someone struggling so desperately with reality.

That's right. All things that travel at the same velocity are the same.
You stick with that.

 

[Seymour Butz]

Wow.

All that just to gloss over the fact that JB won't admit that he missed the context of the original question.

Very telling....

 

[humber]

So how do you establish which one is moving, astronaut or space ship?

What is it with you guys and your fixation with relative velocities?

Two billiard balls colliding at 100km/s, is the same as two oil tankers doing the same. Well, they are both stopped, I suppose.

The astronaut does a push-up against the wall of the spaceship. He is translated 30cm in the process. The energy required to make this move is exactly the same as moving the space station the same distance. The kinetic energy of each will be the same. Mice are elephants.

 

[Dan O.]

The collisions are different! They are different events, not different views of the same event. The result is completely independent of the viewer, all viewers. You can find different combinations that will yield the same final velocities. Do you require proof by exhaustion of calculation?
The final velocities will be what they are for each observer in each case.

The 10kg mass requires more energy to get it to 1/ms, than does a 1kg mass. What is there not to understand?

You need to actually do the math before you can get an answer. To make the problem easier, assume that the astronaut sticks to the space craft when they smush together. That way, you only have to worry about conservation of momentum to solve for motion of the collision result.

You are an observer outside of the spacecraft and to your perspective the spacecraft is not moving. You see the astronaut approaching from your right and crashes directly into the spacecraft. You even were fortunate enough to get the encounter on tape. Now, what difference will you see to prove that spacecraft was at fault for running into the astronaut and not the other way around. The court is going to want to see your calculations for what would be expected to show up on the tape in each case.