Manipulating Momentum

[Tarsier79]

Im not sure Kirk is wanting anything to do with your bicycle as much as you want to his to for spare parts to create your latest pedal powered flowerbowl.

[Kirk]

I don’t see that you mention the concept of throwing the small mass back up to the top. After you get the momentum transferred to the small mass you must throw it.

That momentum is what propels it up

[Kirk]

Maybe we don't have to throw the small mass ...

This thread referred me to investigate water wheels - the Pelton wheel is very interesting. This reminded me of stuff I learned a long time ago when I sold pumps ...

With pumps you have Pressure & Flow. This sort of relates to Current & Voltage in electrical systems. With a fixed energy input - these behave like conserved properties. For example, you can have high pressure by restricting flow, or high flow at much lower pressure.

You can design pumps that are very good at producing flow, but not very good at producing pressure (e.g. axial flow pumps). You can design pumps that are very good at producing pressure, but not very good at producing flow.

From Wikipedia: "Undershot wheels gain no advantage from head. They are most suited to shallow streams in flat country."

This stood out to me - because most mechanisms can be reversed. If we can make a water wheel that is not really affected by 'Head' - could we reverse this and use it to elevate mass, with equal indifference to height?

Water wheels show us that there are two basic mechanisms that can be used to turn a wheel: weight (Force, pressure, Head) OR Momentum. They work quite differently. And you can build different types of wheel, depending on whether you want to exploit a Height differential or a Momentum differential.

Maybe we can use both types within our wheel, and exploit them for our purposes.

Seems to me that there are 4 possible combinations:
gain momentum falling, expend momentum rising
gain force falling, expend force rising
gain momentum falling, expend force rising
gain force falling, expend momentum rising

I'm probably not explaining this very well at all - so think about this:

Whatever mass we let fall, it has to be raised - either Higher or Faster - for energy gain.

If the falling mass rides another mass system - the force of gravity is available for the duration of fall. Pequaide - your Atwoods ideas depend on this.

Alternatively - we can let the mass free fall. What we lose in available Force, we gain in available Momentum at the bottom. This is an option that can be explored.

As far as raising the mass again - we have similar options:
1 - we can apply an Impulse and attempt to 'kick' it higher - with no physical connection, or
2 - we can apply Force, and raise it upwards - as with a jack or elevator - with constant physical connection

I think Kirk is on the right track - exploiting two different reference frames. I think the mass has to be ejected from one reference frame and into another.

2 masses with a difference in mass the lesser one gaining higher velocity from equivalent momentum. It has higher ke though momentum is equivalent. This velocity transformation is essential - doesnt work without it. Thus the transformer (the wheel) is essential. Remember the wheel is a lever but constantly in motion so as an analog it looks like a transformer and a capacitor.

[Kirk]

From what I can tell about the Pelton Wheel - it doesn't work the way you describe it. It seems to me that it works by redirecting linear momentum into angular momentum, and reversing the direction of the momentum vector. That in itself is food for thought ...

Sorry to be the party pooper ... I think you are close, but no cigar.

and when a mass encounters a spring and reverses its vector 180 degrees how is that different than the mass encountering a cup and after traversing the cup has reversed its vector 180 degrees?

[Kirk]

Dont forget the purpose of the wheel is 2 fold - storage and transformation.
I suggest it possess at least 100 times the momentum of the masses being manipulated
My 4 foot wheel was to be cut from a 1/4 inch sheet.

[Kirk]

I think that Kirk is definitely onto something, and furthermore that it could well be almost identical to something I happen to be working on. Kirk, we are talking about teeth on a wheel. We also talking about chains, right ?

no I was using the teeth as magnetic events

[Kirk]

I reread the thread and am of the opinion it still isnt clear enough so here goes:

We are going to drive a 4 foot dia. wheel by dropping 1 pound steel balls at a "target" located 6 inches from the center of rotation. The balls are dropped 1 foot so attain a velocity of 8 feet per second. We are going to spin up the wheel until this target is moving 4 feet per second. An elastic collision imparts 4 units of momentum as relative to the target the ball is only moving 4 feet per second. The elastic bounce results in the ball leaving the target at 4 feet per second but as the target is moving 4 feet per second away from the ball it is moving - relative to an observer looking at the wheel but not on it - at zero, it has come to rest. The target has disappeared in a hole in a "floor" and the ball uses a slight incline to roll to the pickup point for recycling.
Since we want to influence the wheel 1% or less it must normally posesses 800 units of momentum. After the collision it has 808 units of momentum. We then introduce a mass of 1/4 pound to the rim at a place where its flight will be upward to the top of a large wheel that will lower it in a controlled descent to extract energy. Since the rim is moving 4 times the velocity of the inner target a target on the rim strikes the 1/4 pound at 16 feet per second and since it is elastic under zero loss an additional 16 feet per second would be imparted resulting in a total of 32 feet per second which will let the 1/4 pound rise 16 feet.
The momentum of a pound at 8 feet per second is equivalent to a quarter pound at 32 feet per second. We didnt add energy to the system.
We know ke is not a force because it does not obey Newtons 3rd law of motion. Momentum does.
For those having trouble do you see it now?

[nicbordeaux]

Sorry, no I do not see it working. But hope that it does.

[pequaide]

Okay: I see the word “flight� that is good.

Why don’t you let the steel balls pull the wheel into motion instead of bouncing them off of it?

Your wheel is very complex, and I don’t think I agree with your 16 ft per sec plus 16 ft per sec for a total of 32 idea. I think you will only get 16.

I prefer the idea of using all of the wheel's momentum; at least for now.

Most of your ideas are supportable. Until someone comes up with some experimental evidence to the contrary I think we should view Newtonian Physics (F = ma, The Law of Conservation of Momentum; The Three Laws of Motion) as a given. What some see as impossible I view as required for any fundamental understanding of the basic Laws of Physics. These experiments can and have been done with the starting motion of the heavy mass in linear motion; and the final motion of the small mass also in linear motion. With all of the experimental evidence in Newton’s favor; what choice do we have but to believe him one more time.

A 10 kilogram sledge moving 3 m/sec has 30 units of momentum and it can transfer all of its motion to a one kilogram mass. According to Newtonian physics the one kilogram mass MUST be moving 30 meters per second after the motion is transfer. Opponents would contend that it can only be moving (½ * 10 kg * 3 m/sec * 3 m/sec = 45 joules = ½ * 1kg * v * v; so v = 9.48 m/sec) 9.48 m/sec.

Newton would predict that the one kilogram missile would soar over one of your Kansas silos (150.5 feet). Newton’s opponents would predict that it could not make it over a farm house (15.05 feet).

Only those that don’t do these experiments would believe that it could only rise 15 feet. And 15 is the ideal height (all things being perfect: no air resistance no bearing resistance). I will agree that it will be short of 150; but it will be a whole lot higher than 15.

[Kirk]

32 is based on a coefficient of restitution of 1. The literature says tool steel has a coefficient of .97 so is very good. The only thing I found higher than that is the alloy "liquid metal" and I dont think I want to use it as I hear reports of shattering.
Yes, the amount of physics that is published without physical testing or the one I like about a non elastic mass that picks up an equal mass and falls to half velocity but they then give a speech about ke lost to heat and noise. Such rubbish.
When I was in school we had labs and if you were VERY careful you might get within 10% of the textbook value but we see the experiment where the thermometer was dropped from a cliff being spot on. I really have trouble with that. Lots of pencil whipping is done in industry, have seen it.
looks like I shall have to save my pennies for a test to satisfy the nay sayers.

[nicbordeaux]

Im not sure Kirk is wanting anything to do with your bicycle as much as you want to his to for spare parts to create your latest pedal powered flowerbowl.

Tarisier, do you have any idea of the potential energy contained in a big flowerbowl ? Total release (no flower bowl left, not a bit of juice then a depleted flower bowl) and you could produce more energy than all the oil in the world.

[greendoor]

We are going to drive a 4 foot dia. wheel by dropping 1 pound steel balls at a "target" located 6 inches from the center of rotation. The balls are dropped 1 foot so attain a velocity of 8 feet per second. We are going to spin up the wheel until this target is moving 4 feet per second. An elastic collision imparts 4 units of momentum as relative to the target the ball is only moving 4 feet per second. The elastic bounce results in the ball leaving the target at 4 feet per second but as the target is moving 4 feet per second away from the ball it is moving - relative to an observer looking at the wheel but not on it - at zero, it has come to rest. The target has disappeared in a hole in a "floor" and the ball uses a slight incline to roll to the pickup point for recycling.
Since we want to influence the wheel 1% or less it must normally posesses 800 units of momentum. After the collision it has 808 units of momentum. We then introduce a mass of 1/4 pound to the rim at a place where its flight will be upward to the top of a large wheel that will lower it in a controlled descent to extract energy. Since the rim is moving 4 times the velocity of the inner target a target on the rim strikes the 1/4 pound at 16 feet per second and since it is elastic under zero loss an additional 16 feet per second would be imparted resulting in a total of 32 feet per second which will let the 1/4 pound rise 16 feet.
The momentum of a pound at 8 feet per second is equivalent to a quarter pound at 32 feet per second. We didnt add energy to the system.
We know ke is not a force because it does not obey Newtons 3rd law of motion. Momentum does.
For those having trouble do you see it now?

Sorry - i'm fairly thick and have trouble picturing things. (I get swamped with a million possible solutions, and I can fail to see what you have in your mind).

You describe a ball dropping at 8 fps impacting a wheel target that is moving at 4 fps. I assume you mean these vectors are in the same direction (more or less, comparing linear with angular, which I am very comfortable with).

What makes you think the ball will be seen to "come to rest" as observed from outside the wheel, but not from the wheel? I don't get it.

Are you assuming that in a collision, both masses end up accelerating at the same velocity? Because that's not how it works - not unless they have the same mass. Are you engineering the wheel to have the same mass as this ball?

You seem to be working on the assumption that by hitting the wheel when the wheel is travel away at half speed, that this somehow transfers all momentum to the wheel ... I don't think it does.

By having the wheel traveling away from the ball, it greatly reduces the relative speed difference, so the impact is much less than if the wheel was stationary. If the wheel was traveling at full speed, there would be no impact at all.

So I very much question whether the ball would bounce back at 4 fps or not, and whether this imparts as much momentum to the wheel as you think it might ...

Or have I missed something?

[greendoor]

I also don't get why you are launching 1/4 pound up 16 feet. You dropped 1 pound - so are you going to launch four of these 1/4 pound weights up 16 feet? It seems there isn't enough momentum to do this four times for the one drop of the 1 pound mass ...

Or have I missed something?

A picture is worth 1000 words - and even then I probably won't get it ...

[Kirk]

One pound dropped 1 foot launches one 1/4 pound mass 16 feet - their momentum is equivalent. The ke is 400% dfferent

[Kirk]

[quote="greendoorSorry - i'm fairly thick and have trouble picturing things. (I get swamped with a million possible solutions, and I can fail to see what you have in your mind).

You describe a ball dropping at 8 fps impacting a wheel target that is moving at 4 fps. I assume you mean these vectors are in the same direction (more or less, comparing linear with angular, which I am very comfortable with).

What makes you think the ball will be seen to "come to rest" as observed from outside the wheel, but not from the wheel? I don't get it.

Are you assuming that in a collision, both masses end up accelerating at the same velocity? Because that's not how it works - not unless they have the same mass. Are you engineering the wheel to have the same mass as this ball?

You seem to be working on the assumption that by hitting the wheel when the wheel is travel away at half speed, that this somehow transfers all momentum to the wheel ... I don't think it does.

By having the wheel traveling away from the ball, it greatly reduces the relative speed difference, so the impact is much less than if the wheel was stationary. If the wheel was traveling at full speed, there would be no impact at all.

So I very much question whether the ball would bounce back at 4 fps or not, and whether this imparts as much momentum to the wheel as you think it might ...

Or have I missed something?[/quote]
yes the ball and the wheel have the same vector
relative to an observer outside the wheel the ball is moving at twice the velocity of the wheel. The 4 feet per second of the wheel is subtracted from the 8 of the ball because the difference is 4. So the ball collides with the wheel at 4 feet per second. It is an elastic collision so it rebounds at 4 feet per second relative to the wheel. To an observer outside the wheel the motion of the ball is now zero - all the momentum of 8 feet per second has been transferred.
No -the wheel is much larger than the ball. Did you miss the post where I said its momentum should be at least 100 times larger?
Re transfer of momentum guarantee it does- talk to a hydraulic engineer. Pelton wheels have been around for a long time.
They are quite effficient.