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### Momentum Inversion Calculation

07/11/2012 9:39 AM

Calculation for moment inversion

Dear all, when I was drafted to the IDF, 1956, I became a sherman tank driver.

I was very frustrated because I couldn't understand why a 35 tons tank, equipped with a 380 hp engine isn't able to go faster than 32 km/h, and the Tank carrier, which is a truck with 200hp, will go with 6o tons, about 60 km/h?!

Until some friend told me that the tank's engine has to overcome the moment inversion of the tank's tracks.

This means that tracks move first from the back to the front, there their direction is inverted from the front to the back.

The faster the tracks move, their momentum is higher, and this requires more energy.

According the formula for kinetic energy, this should be calculated in the following way:1/2mv^2

Where I think that m in this case isn't the track's mass, but the track's moment, this means its speed[m' per second] times the mass in k.g.

Then it will be used as the m in the formula ½ m v^2.

I need it for some practical use, when I'll end this task I'll tell you the whole story.

So, am I right or wrong in my calculations?

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#1

### Re: momentum inversion calculation

07/11/2012 9:52 AM

I would say environmental conditions dictates the ratios and power transmission requirements and how they are implemented.

The tank carrier is just that, which hopefully uses something that resembles a road, while a Sheman, well, can hits the hedges and rough terrian....

So inversely, you can say the tank carrier can't go where the Sherman can.

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#2

### Re: momentum inversion calculation

07/11/2012 10:06 AM

Dear Phoenix: If you want to shoot- shoot!

If you think that my calculation are wrong-say it!

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#3

### Re: momentum inversion calculation

07/11/2012 10:15 AM

What calculation?

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#4

### Re: momentum inversion calculation

07/11/2012 10:25 AM

maybe he mean't formula?

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#7

### Re: momentum inversion calculation

07/11/2012 10:42 AM

The formula was developed by scientist, I have to calculate which formula to use!

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#5

### Re: momentum inversion calculation

07/11/2012 10:31 AM

Is more involved than a simple formula.

for one, Your have to start with the track effiecency. And Why track. And thats just the start.

You were doing two comparision of two total different vehicles with different task functions.

As an example comparision, File and a hand saw. Why does the hand saw cut wood so much faster then a file?

BANG! sorry, I did'nt know it was loaded

That was the point I was getting at. I think it would be simplier for you to talk to your friend first about Moment Inversion and how it applies.

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#6

### Re: Momentum Inversion Calculation

07/11/2012 10:39 AM

I am inclined to believe the tank's engine must overcome, gee I don't know, about 400 sets of bearings. Compared to a wheeled transport vehicle there are a LOT of spinning drivetrain parts in a track vehicle.

While the tank was in motion, did you ever disengage the drive and let it coast along?

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#8

### Re: Momentum Inversion Calculation

07/11/2012 10:45 AM

Not to mention the drag for turning 35 tons.

I believe he's concentrating on a different matter.

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#9

### Re: Momentum Inversion Calculation

07/11/2012 2:10 PM

I'm afraid your friend was guessing.

He reasoned that the track running along the top of the wheels was going forward at twice the speed of the tank, whilst the track in contact with the ground was stationary. And he figured that to keep accelerating and decelerating all that mass would take a lot of energy. However the same would be true to a lesser extent for an ordinary wheel, but, try this mental experiment: roll a wheel on an axle along a table and then just raise it off the table but keep the same horizontal speed; the the top of the wheel will keep going at twice the speed of the axle and the bottom of the wheel will remain stationary with respect to the table.

(Which part of a train always goes backwards? The part of the flange beneath the rail.)

The tank track would also continue to roll in exactly the same way if there wasn't a huge amount of friction to overcome.

There are some mechanisms based on flattened rolling hoops of spring steel which are very efficient.

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#10

### Re: Momentum Inversion Calculation

07/11/2012 3:23 PM

Just change the gear ratios in the drive system and I'll bet you can do 50 km/h or even more, until you throw a track.

Oh, and let's hope you don't have to stop on a dime.

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#11

### Re: Momentum Inversion Calculation

07/11/2012 3:58 PM

Dear Lyn- It seems to me that you never drove a sherman tank, that had only 12 hp per ton. Modern tanks have about 20 hp per ton. If you would drive any tank, then you would discover, that it was very rarely for the engine to reach the maximum rpm 2400.

I really was disapointed to see that you didn't relate my question!

Just tell me if I'm wrong!

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#12

### Re: Momentum Inversion Calculation

07/11/2012 4:09 PM

Az native, I don't know about the others but I am very curious:

"...when I'll end this task I'll tell you the whole story."

Have you picked up a surplus tank and now you're fixing it up, trying to get some extra oomph out of her?

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#13

### Re: Momentum Inversion Calculation

07/11/2012 4:26 PM

HP to weight ratios,

What that doesn't tell is the efficiencies of the drives.......... The drive could really be inefficient with that being cumbersome and the like will bring down your HP per weight ratio.

And the drive would include clutching, transmission, ...........

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#14

### Re: Momentum Inversion Calculation

07/11/2012 5:19 PM

Of course I've never driven a Sherman Tank. I'm not that old.

I have driven an M-60A3 Main Battle Tank during tests at Edwards Air Force base. Does that count?

I still can't relate to your question. Sorry.

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#32

### Re: Momentum Inversion Calculation

07/15/2012 11:51 PM

Is the 2400 RPM limit for the engine loaded to its rated HP output or it in a no load high idle speed mode?

Any engine with a sloppy or relaxed governor system will have two very different running RPM's relating to its loaded VS unloaded state.

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#33

### Re: Momentum Inversion Calculation

07/16/2012 12:43 AM

Dear friend-Last time I drove a sherman tank with the R975 continental engine was 1970, I'm afraid that I'm not able to answer your question, but I can tell you that when the road went down ,then we could hear the governor comming into action and the engine was growling with ups and downs!

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#34

### Re: Momentum Inversion Calculation

07/16/2012 3:36 AM

Well theres your problem! The engine governor was what was limiting the overall speed.

As far as track momentum and speed reference the ground is not the correct reference point. The machine is.

From the perspective of the machine, being its the driving power source for the system, the top of the track is moving exactly the same speed and with exactly the same magnitude of forces as the bottom but in exactly the opposite direction thus they cancel each other out. Simple force vector analysis when done with the machine as the reference point will give you a net value of zero when the two track halves are added together. The link on the top going over the front end roller and down has the exact same energy as the bottom link going under the rear roller and up but at exactly the opposite direction thus their perspective momentum energies cancel each other out. Neither gains or looses.

The only energy losses in the tracks are from friction between all the links, bushings, pins, and rollers. If they are treated mathematically as perfect frictionless bearings the only energy involved is that which get the total mass of the track assembly's up to what ever speed you choose. After that they are frictionless and require zero additional energy input to keep their overall velocity constant.

As far as the scientist who did what ever all I know is I had a math instructor in college who did a lot of stuff like that as well. When it came to the obvious principals of applied science and engineering he was probably one of the most brilliantly clueless/useless dumbass's I have yet to meet so far.

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#35

### Re: Momentum Inversion Calculation

07/16/2012 4:19 AM

I think that better you should do what I have done- after I got the idea that the fault of the moped that I want to consruct might be the inversion of the momentum, I have done 2 things:

1]I calculated using the formula, and the data, the results suited well to what I saw.

2]I discussed the topic with a friend of mine, who is professor for Aerodynamic, [if I could reach him earlier I wouldn't bother all you cr4 readers, but he was sick]

The professor confirmed my calculations.

Any more questions?

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#15

### Re: Momentum Inversion Calculation

07/12/2012 2:14 AM

Dear friends-I introduced here a question under the title of "Momentum inversion calculation" but none of you- the people who bothered themselves to answer me, related my question: if I'm right with my calculation. I marked well the formula that I used.

Now I have another question: What was wrong in the way that I introduced my question?

First I told the story that brought to be acquainted to the " momentum inversion", I thought that the story will do it little "juicy", but it seems that the people who answered, related only to the story, and abandoned my question with the formula.

So I have to asume that something in the way that I introduced my question was wrong!

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#16

### Re: Momentum Inversion Calculation

07/12/2012 3:01 AM

It sounds like hogwash to me!

The tank is geared to travel at 32km/h over rough terrain, if it travelled on a road it could be easily re-geared to travel as fast as the carrier.

There are tanks which can travel at high speeds on a road, but that is not their main purpose in life! They are designed to go, under control, where wheeled vehicles cannot.....

Horses for courses....

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#17

### Re: Momentum Inversion Calculation

07/12/2012 3:45 AM

Even on a flat asphalt road the sherman tank was unable to go faster than 35 km/h!

Why don't you answer my question?!

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#18

### Re: Momentum Inversion Calculation

07/12/2012 4:10 AM

I DID - the engine has a maximum rpm, the gearing and maximum rpm set the maximum speed of the tank, if the gearing was changed (higher) it could have travelled considerably faster on the road, but then because the gearbox has a limited number of ratios, that would have compromised its ability in rough terrain where it needed the very low gears ...... The Low gears on the tank will probably be lower than those on the truck, the truck also has very low gears, but it probably has 16 or more forward gears to be able to get up to 60+kph ..... how many forward gears did the sherman have?

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#19

### Re: Momentum Inversion Calculation

07/12/2012 4:26 AM

My question was about the caculations and the formula of moment inversion!

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#20

### Re: Momentum Inversion Calculation

07/12/2012 5:14 AM

You are correct that momentum = mv

and kinetic energy = ½mv²

But, what we have all tried to point out is that the rest of your assumptions about the reason why the tank does not go as fast as the truck are incorrect.

So if you're basing your later work on incorrect assumptions, we all reasoned that there's no point in the "later work".

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#21

### Re: Momentum Inversion Calculation

07/12/2012 5:21 AM

No great mystery...

1. The 200hp truck has minimum contact with the road surface (compared to the tank), while the wheels run on free bearings, and the gearing is calculated for a load dictated by the bed/ chassis strength, and the capacity of the truck's suspension.

2. The tank's tracks do not push/ pull. The tank's actual motion occurs inside the tracks .....like a railway locomotive laying track continuously in front of itself. You will have noticed there is no 'tread' on the tracks, and this is what makes the system work so well. The tank is running on a continuous smooth steel 'road'.

3. The tank is necessarily an all-terrain vehicle. This means there must be free-play in the track length, and all of the suspension members along the length of the track need to be somewhat independent of each other. It is within this design that the maximum speed will be governed, else the tracks may shuck like a bicycle chain.

I don't dispute the fact of 'moment inversion' especially if there are both front and rear drive axles...I just don't think that that the concept governs top speed.

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#22

### Re: Momentum Inversion Calculation

07/12/2012 8:47 AM

You are wrong. The letter m in your equation for kinetic energy stands for mass, not for mass times speed.

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#23

### Re: Momentum Inversion Calculation

07/12/2012 9:10 AM

"This means that tracks move first from the back to the front, there their direction is inverted from the front to the back"

The track never moves front to back. Once it makes contact with the ground it stops and it's the tank that moves forward over it. As Hilton pointed out the track is continuously laid in front of the tank then picked up again at the back.

There is no "moment inversion" that you speak of.

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#24

### Re: Momentum Inversion Calculation

07/12/2012 9:17 AM

I think your point would be correct if the track would be infinite, but the track is moving below and above the tank's wheels!

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#25

### Re: Momentum Inversion Calculation

07/12/2012 9:25 AM

"... but the track is moving below and above the tank's wheels!" And at the front and back ends there is a direction change, sort of like this:

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#26

### Re: Momentum Inversion Calculation

07/12/2012 9:26 AM

To make my pointe clearer:

Leave now the track, think about some phisycal body, it's mass is m, now accelerate it from left to right at speed v, it's moment will be mv, okay?

Now figure what will be the force needed to pull it [or push] 120 degrees aside?

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#29

### Re: Momentum Inversion Calculation

07/12/2012 10:54 AM

You are confusing your terminology. mv is MOMENTUM, not moment.

If all you really want to do is determine how much force is required to change the direction of a moving object, then a simple force vector diagram can give you the answer a whole lot easier than all the calculus in the world. Moments have noting to do with it.

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#30

### Re: Momentum Inversion Calculation

07/12/2012 11:51 AM

120 degrees??...Is this what you are refering to (see below).

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#27

### Re: Momentum Inversion Calculation

07/12/2012 10:44 AM

I've never heard the term "moment inversion".

Understanding the concept of what you're asking is, IMO, easiest to understand if you define a coordinate system (origin in the "center" of the tracks) that moves WITH the tank.

In this coordinate system the tracks have a generalized circular motion, thus a moment of inertia and angular velocity. Standard circular motion equations apply to find the power, energy, etc for just the tracks.

This is just for the spinning tracks and must be added to the results from the standard kinetics (where the coordinate system is earth referenced).

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#28

### Re: Momentum Inversion Calculation

07/12/2012 10:52 AM

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#31

### Re: Momentum Inversion Calculation

07/12/2012 3:10 PM

First top spead is a function of arodynamic drag and friction losses. The power to speed is a cubic relation above about 15 MPH. That is:

HP = aconstant * speed^3

where HP is power available at the driving contact point. Tire or tread to ground.

The engine HP is reduced by friction losses through the drive train and other driven equipment such as generators etc. HP at the drive contact is reduced.

Momentum is more closely related to acceleration sense accelerating is changing momentum. Looking at Newton's basic laws of motion:

force = mass * acceleration. or acceleration = force/mass,

a = F/m

Intergrating acceleration gives you velocity (also commonly called speed)

v = a * t + v0

Intergrating again gives the distance traveled:

d = 1/2 * a * t^2 + t*v0 + d0

The momentum formula 1/2 * m*v^2 is changing with acceleration:

1/2 * m* (a*t+v0)^2) = 1/2 * m*(F/m * t + v0)^2

Momentum is a form of energy. All moving mass has momentum and there is an energy transfer with any change in momentum. The tank tracks would of course have momentum and would of course reduce the accelerating force. That would effec the acceleration and deacceleration of the vehical but little or no effect on it's top speed. Other then friction.

The momentum of the track is trying to go in a stright line and takes an oposing force to make it conform to it's constraned path. I would say the top speed limit is a design issue. Probable by design to keep the forces on the track and it constrants within design limits.

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#36

### Re: Momentum Inversion Calculation

07/16/2012 5:02 AM

So if you get enough people to use the wrong formula that makes a difference in the results?

You lost me. How does an aerodynamics person fit into a misunderstanding of mechanics problem on a tank?

When did the moped come into this tank track issue? Did I miss something along the way?

When you get a chance ask your aerodynamics person if he uses the speed of the wing through the air as the reference to the air speed or the ground as the reference point in his calculations. I bet if he uses one instead of the other he gets the wrong answer and his plane falls out of the sky!

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#37

### Re: Momentum Inversion Calculation

07/16/2012 6:33 AM

Dear friend- I have no idea about the education in North Dacota, but I know that in Israel, and most likely in most of the countries, every aerodynamic engineer is first a mechanical engineer, the laws of physics are equal!

That doesn't mean necessarily that any mechanical engineer is an aerodynamic engineer!

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#38

### Re: Momentum Inversion Calculation

07/16/2012 2:38 PM

Do you have a valid link to something that accurately describes 'Momentum inversion' in physics?

I did all the digging I could do and so far the term 'momentum inversion' is related to stock market and similar financial investment ( http://www.querycat.com/question/44516d8f880aa77e9460e179de68ae5e ) and nothing physics related that I could find.

But then that math professor I was rather good at applying incorrect math formulas to the wrong applications as well.

From what I can recall of my physics classes a moving track is a combination of rotational kinetic energy, angular momentum, and linear momentum moments of multiple objects all working together as one large object, these formulas seem familiar http://physics.bu.edu/~duffy/py105/AngularMo.html , where the laws of conservation of energy still apply.

(unless you use the wrong reference point for your speeds, directions and energies. overunity theory principals 101.)

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#40

### Re: Momentum Inversion Calculation

07/16/2012 3:07 PM

I could not find it anywhere, closest I found in my dictionary states this.

[mathematics] Given a point O lying in a plane or in space, a mapping of the plane or of space, excluding the point O, into itself in which every point is mapped into its inverse point with respect to a circle or sphere centered at O.

The interchange of two adjacent members of a sequence.

[mechanical engineering] The conversion of basic four-bar linkages to special motion linkages, such as parallelogram linkage, slider-crank mechanism, and slow-motion mechanism by successively holding fast, as ground link, members of a specific linkage (as drag link).

Parker, Sybil P.,(ED) (1994), 'McGraw-Hill Dictionary of Scientific and Technical', (5TH) Edition, McGraw-Hill, New York, NY, pg 1046

.....I also found it on this web site and pasted it, the link is here .....

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#41

### Re: Momentum Inversion Calculation

07/16/2012 3:26 PM

Yes-you are right regarding that me too couldn't find any thing in the internet.

But why not use the common sense? BTW- math isn't my cup of tea!...

Just try to consider the forces:We know that the track has some momentum, do you think that we can change the direction of this momentum without any "penalty"?

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#76

### Re: Momentum Inversion Calculation

07/23/2012 9:14 AM

Obviously there are some that no nothing about aerodynamics of
automobile design. Aerodynamics is primary limiting factor of the
top speed of a vehicle.

The point being that even if there were such a thing as Momentum
Inversion it would have nothing to do with the top speed of a tank.

The top speed is limited by the power you get to the ground
propelling the vehicle. That power has to overcome friction and
aerodynamic drag.

The power to do 60 MPH is 8 times the power required at 30 MPH
and at 90 it is 27 times the power required at 30.

And I am not an aircraft engineer. I am more into automobile
design.

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#77

### Re: Momentum Inversion Calculation

07/23/2012 9:27 AM

Interesting, seems to be a pattern, or is that a rule of thumb.

The pattern that comes to mind is the inverse square.

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#39

### Re: Momentum Inversion Calculation

07/16/2012 2:51 PM

Consider these two tank tracks: imagine them both being driven along at the same speed.

Do you think that the long one would suffer more from your friends hypothetical momentum inversion problem?

Consider links at positions A and positions B. Do you believe that they "continue in their states of rest, or of uniform motion in a straight line"; without any forces acting on them?

Do you think that links at positions C1 and C2 will experience different forces on them? Likewise the links at D1 and D2. Would it make any difference if the tanks motions were from left to right or right to left?

Although there would be considerable mechanical difficulties to do it in practise you could continue to move the two "end" wheels together until you had just a simple wheel.

Does that make sense?

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#43

### Re: Momentum Inversion Calculation

07/17/2012 3:58 AM

Did you think about the probability that the longer is the chain/track, the bigger the mass?[mass times speed=momentum]

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#44

### Re: Momentum Inversion Calculation

07/17/2012 5:12 AM

Yes I did. I think that point was covered by Newton's first law

Consider links at positions A and positions B. Do you believe that they "continue in their states of rest, or of uniform motion in a straight line"; without any forces acting on them?

The only place where momentum changes (if the tank is maintaining a constant velocity) is at the ends, and, the situation is the same for the long track the, the short track or a simple wheel.

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#45

### Re: Momentum Inversion Calculation

07/17/2012 9:13 AM

My laziness brings to try to concentrate on wheels:

I think that the best and most accurate definition for a wheel is a rotating body that is well balanced.

I think that once you accept this definition, you'll be able to accept my approach for the momentum inversion.

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#50

### Re: Momentum Inversion Calculation

07/17/2012 11:33 AM

I'm afraid not.

Go back to my post #39 and answer the questions one at a time.

If I'd been more careful about creating the tracks I could have made the long track contain exactly twice as many links as the short one. If I'd done that then each individual link in the short track would have its momentum reversed twice as often as the links in the long track, but, don't forget there are only half as many, so, the total amount of "momentum reversal" would be the same.

Again I could have made the short track so that it had exactly twice as many links as a "wheel". The same argument would apply.

In the case of the tracks the links which are travelling in a straight line at constant speed are not "costing" you anything. In all three cases the links which are in circular motion are equivalent. They do not "cost" you anything because the acceleration is towards the centre of the arc.

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#72

### Re: Momentum Inversion Calculation

07/20/2012 4:39 AM

I know that there has been a little sarcasm here, but, believe me we are trying to help you understand.

I don't think laziness will prevent you from answering the simple questions one at a time.

1.) Do you think that the long track "suffers" from this problem of momentum inversion more than the short track? Yes or No?

2.) Do you believe that the links which are stationary or moving forward in a straight line at twice the speed of the tank obey Newton's first law? Yes or No?

3.) Thinking about one of the links which has just started to move around the "wheel" at the end. Do you think that the total forces on it are different dependent on whether the track is 5 metres long or 2 metres long? Yes or No?

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#73

### Re: Momentum Inversion Calculation

07/20/2012 10:07 AM

Not to derail your line of questions here, but I think it is important to clear a couple things up for the lay-person who might be reading this.

1) Your statements about forces regardless of track length are quite correct (given a constant velocity).

2) it is equally important to note that a longer track (having more mass) will require more energy to change velocity.

I have a feeling this is where the confusion is.

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#74

### Re: Momentum Inversion Calculation

07/20/2012 1:21 PM

I'm pretty sure the OP is convinced that "the problem" exists for a tank moving in a straight line at constant speed.

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#75

### Re: Momentum Inversion Calculation

07/21/2012 3:57 AM

This morning while I was hiking I was thinking about your questions, and I started to calculate, and to my astonishment I found something interesting:

Let us assume the tank's mass is 30 tons.

Both track's mass:2000 kg.

The tank's speed - 36 km/h - which is 10 m' per second.

So the tank's momentum will be 300000 kg.

The track's momentum- 20000 kg.

According the inversion of momentum formula that is written nowhere the energy requirement is 1/2*20000*10^2=1000000 w!1000 kw....

How could it be I asked myself?

And here comes the interesting answer:

The tank's momentum! The faster the tank is moving, he has a bigger momentum, in case of a good road there is less friction -[in the case of tracks, stones on the road will be considered as "friction"],so the tank's momentum will compensate for the energy loss of momentum inversion!

And this will enable to reach the 36 km/h

How do you like this idea?

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#42

### Re: Momentum Inversion Calculation

07/16/2012 7:51 PM

Finally. I think I may be getting an insight into this tortuous proposition. I think it is about changing the direction of travel of the vehicle.

Think of the gyroscope! the valued property of the gyroscope is that moving the axis of rotation is difficult because of the energy in the wheel. The same energy is present in the rotating wheels of vehicles, pedaled or powered. There is a considerable amount of energy in the tank tracks, to the point that they usually apply the brakes to one track in order to turn.

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#46

### Re: Momentum Inversion Calculation

07/17/2012 9:44 AM

I searched for Momentum Inversion through all my engineering books.

· Standard Handbook for Mechanical Engineers-Marks,

· Standards Handbook of Engineering Calculations-Hicks,

· Machinery's Hand Book - 22nd Edition,

· Basic Graphical Kinematics-Kepler,

· Engineering Mathematics Handbook-Tume-Walsh,

· Applied Physics- Tippens,

· Standard Handbook of Machine Design- Shigley-Mischke

· Machine Design Data Handbook - Lingaiah

And 4 others and absolutely no mention of Momentum Inversion.

Has anybody even heard of this?

Does it go by another name?

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#47

### Re: Momentum Inversion Calculation

07/17/2012 10:31 AM

I've never heard the term either.

I still maintain that the equations to describe the motion, energy, etc of the tracks are found in a first year dynamics (the course most of us took our second semester of school) book in the circular motion section.

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#48

### Re: Momentum Inversion Calculation

07/17/2012 10:37 AM

I must have been sick that day we covered it......

I'm just surprised there were no references about it, makes me think its called by another name..........or something that was started by speculation and labeled with an important sounding name.

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#49

### Re: Momentum Inversion Calculation

07/17/2012 10:58 AM

My guess is that it is one of those situations that someone doesn't/didn't ever learn circular dynamics and tried to reinvent the wheel (ha!). It wouldn't be the first time.

It is a great sounding name though.

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#51

### Re: Momentum Inversion Calculation

07/17/2012 2:30 PM

So why doesn't this momentum inversion issue cause problems with efficiency losses mechanical drives systems like chain drives and belt drives?

In many applications there are massive belts and chains being operated at high speeds and they are not considered inefficient by any means.

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#64

### Re: Momentum Inversion Calculation

07/18/2012 8:08 AM

The reason for my op, was that there is enormous power loss in the drive train for my home build moped, I designed it with the help of some engineers, and we didn't know why!

Untill I came up with the idea that the Honda engine works on 7000 rpm, and hence driving the chain 14 m' per sec, causes lot of power loss!

I put al this data, including the formula on the OP,but none of the guy here related it!

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#66

### Re: Momentum Inversion Calculation

07/18/2012 8:53 AM

So your saying, the honda is more efficent at a higher RPM

Velocity and output power are inversely tied together

Higher the velocity the less HP output (less efficent)

Lower the velcity more HP output

And yes, I for one could not answer it. Because I am not really sure what you are asking.

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#67

### Re: Momentum Inversion Calculation

07/18/2012 9:01 AM

"Untill I came up with the idea that the Honda engine works on 7000 rpm, and hence driving the chain 14 m' per sec, causes lot of power loss!"

14 meters per second? 14 METERS per SECOND?

Great Scott!

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#69

### Re: Momentum Inversion Calculation

07/18/2012 8:29 PM

So for a continuation on this what is your engines output power rated at and how heavy was the scooter plus rider?

Reason being I am sure more than just a few people here can work out the estimated power it would take to move X amount of mass at 14 meters a second and if your engine falls short.... well.

As far as 14 meters/second goes most chainsaws run at speeds higher than that with little bitty engines that often don't top 2 - 3 HP peak! So how are they getting that high of chain speeds plus enough power left over to cut the wood as well?

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#52

### Re: Momentum Inversion Calculation

07/17/2012 3:22 PM

You guys are trying to figure out something that doesn't exist. There is no such thing as "moment inversion" calculations in tank design. The OP took two technical sounding words and stuck them together to create a made up theory. It's bogus...and if it isn't bogus the he should prove it to us by posting a pic of the written document where he claims his "engineering friend"..."confirmed his calculations".

Tanks move slow with a given HP engine compared to a wheeled vehicle because a tank track and it's suspension system are ridiculously inefficient compared to a wheel !! End of story. As other posters have stated...think of all the friction you need to overcome in the entire system (transmission, suspension, sprockets, etc). You need higher reduction gearboxes per given HP engine to provide enough torque to overcome these inefficiencies which inevitable slows the forward movement.

Anyone with an intuitive mechanical sense can see this.

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#53

### Re: Momentum Inversion Calculation

07/17/2012 3:24 PM

You guys are trying to figure out something that doesn't exist. There is no such thing as "moment inversion" calculations in tank design. The OP took two technical sounding words and stuck them together to create a made up theory.

We already came to that conclusion.... #46 and #48

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#54

### Re: Momentum Inversion Calculation

07/17/2012 3:43 PM

I didn't read all the posts. I couldn't be bothered after I asked a question (#30) and got no response. It's good to know I'm not the only one with common sense.

Cheers.

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#55

### Re: Momentum Inversion Calculation

07/17/2012 3:50 PM

I was going to remove it, but never did, because I was curious to see what that meant, but it just dragged on with specualtion........ This morning, I just rifled through my engineering books I had on hand and nothing really close.

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#57

### Re: Momentum Inversion Calculation

07/17/2012 4:03 PM

I just noticed someone gave my question about the 120deg (#30) a good answer score!...it's a question not an answer.

I guess I should have made the "??" bold.

Cracks me up.

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#58

### Re: Momentum Inversion Calculation

07/17/2012 4:09 PM

look at the responce........ or maybe it meant GQ .......good question

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#56

### Re: Momentum Inversion Calculation

07/17/2012 4:00 PM

Earlier than post 46.

I was suspicious at post 34 and as I read it others were suspicious well before me and I have the impression from your own posts from 1 up to 13 you already had the most plausible causes already nailed down.

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#60

### Re: Momentum Inversion Calculation

07/17/2012 4:12 PM

ok, you get the hound dog award for the first at 34,

as far as from me, my earlier posts was really............. speculation, just to keep it alive, I was hoping someone would have come up with something so I may learn something............

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#61

### Re: Momentum Inversion Calculation

07/17/2012 7:41 PM

"I was hoping someone would have come up with something so I may learn something............"

Life gets a bit more boring when you start having to face the plausibility that just maybe you really do know it all huh?

Personally I was just looking for a good educational argument myself especially when the 'If you ever observed...' bit came up. Trust me if its mechanical I have probably observed and pondered on it far more than what may be considered healthy!

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#59

### Re: Momentum Inversion Calculation

07/17/2012 4:10 PM

I just noticed Doorman's last sentence in #6

"While the tank was in motion, did you ever disengage the drive and let it coast along?"

Exactly!...GA from me.

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#62

### Re: Momentum Inversion Calculation

07/18/2012 2:49 AM

Listen young man, I'm "hyper creative" but I wouldn't dare to create such a theory!

Me too-I was really disapointed to find that I can't find nothing about "momentum inversion", but the commig days will show us more evidence.

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#63

### Re: Momentum Inversion Calculation

07/18/2012 7:25 AM

That sounds more like a statement of faith than a scientific judgement.

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#65

### Re: Momentum Inversion Calculation

07/18/2012 8:36 AM

Wonderful. I sit here in eager anticipation. By the way, would you be so kind to let us know what you meant by "120 degrees aside" in your previous post #26. I'm losing sleep not knowing what you were referring to. And I drew a pretty picture too! Cheers.

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#68

### Re: Momentum Inversion Calculation

07/18/2012 7:34 PM

By the way TerraMan, in your previous post #30 where the drawing says 'Track moving backward and downward, actually the track is moving forward - consider the horizontal position of a cleat as it comes over the top guide roller and where it finally contacts the ground.

In fact the only time a tank track cleat will move backwards is if the track is slipping on the ground, which can happen, but without slippage on the ground all parts of the track are either stationary(on the ground) or moving forward (off the ground).

The rate of forward movement of each cleat changes as it accelerates from stationary to twice vehicle speed at the rear and decelerates from twice vehicle speed to stationary at the front.

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#70

### Re: Momentum Inversion Calculation

07/18/2012 9:20 PM

Yes I know it never moves backwards. I stated that in my 1st post #23. OP never understood it so in #30 I asked the question what he meant by 120 deg and did he interpret "backward" (or in his words "...front to back") as the way I illustrated it. I should have written next to the backward arrow "...do you think the tracks are moving backward here?"

Your explanation is clearer...but I still don't think he gets it!

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#71

### Re: Momentum Inversion Calculation

07/18/2012 9:48 PM

I agree, it is pretty clear to the rest of us!

Gearing, mechanical losses and top working engine rpm sums it up