lposter

THis has been annoying me for most of the day and Iam therefore seeking help.

Lets say a tank has two tracks and each weighs (on their own) 1 kg. Lets say the tank alone without tracks has a weight of 5 kg.

If we take the tank with the two tracksmounted .... it would seem to me that it has two weights? The first being 7 kg and that being the one that you would see if the tank was dumped on abathroom scales.

The second being the effective weight the tank has to move around which is I assume somewhat less? As around half the weight of each track isnt being supported by the tank ? ie. the tank is sitting upon about half the track and this cannot contribute to the effective weight of the tank? In which case the effective weight of the tank is more like 6 kg and not 7?

Is this correct or not? ie. that a measure of the mass of the trackless tank plus the two tracks is not an accurate measure of the actual mass thant the power unit has to move around.

Any input?

p

FreakyDude

kind of yes and no on this. the weight is the weight. The Mass changes to a degree because you don't count most of the weight of the tracks when the tank is static BUT thinking this it means the mass increases when the tank starts moving as you now have to include the weight of the tracks in its motion. The tracks are not weightless if you stop the tank. Remember inertia will play on this.

heavyaslead

That is essentially correct as you stated.

However physics begs to consider intertial weight and drag as factors figuring the power required to move around.

For instance of the same weight, wide tracks have significant resistance to the transmission when turning as opposed to narrow tracks.

Also the terrain type and gousser style affect the effective power movement not withstanding the weight of the track alone.

For instance, Track pads will offer less resistance to high speed movement (like the M1A1) compared to teethed tracks.

lposter

This really isnt as simple as Ihoped it would be.

I was trying to calculate battery capacities/lifetimes etc.

In those calculations it involves the mass of the tank and plus factors such as slope/inertia etc etc etc.

And doing them.....I reasoned that if the calculations include factors like angles, ground roughness factors, etc etc, then the effect of the track lying on the ground must have been therefore accounted for (as its that piece of the rack things like friction are acting on).

In which case Ithen reasoned that the "mass" of the tank to be used in calculations must therefore be the mass of the tank not affacted by such forces (ie. the mass of the tank minus the mass of the tracks lying on the ground).

Otherwise I would be overestimating the force needed to move th etank up th ehill and thereby underestimating the drain on the battery.....

Maybe its just easiest to accept that underestimation and live with it.......

p

Green Amphibian

It is much better to have more power available to use than not enough. Good enough for me anyway.

Herman

heavyaslead

Remember too, the gear ratio will 'even out' the inertial draw on the system. The motor curve is usually most efficient at rated voltage, so actually power draw can go down with greater speed.

Motor curves or torque curves are the first stop when considering the efficient use of battery power.

How the motor is run is perhaps more a determining factor on battery life than mass.
Motors have a huge (relatively speaking) power draw when starting and multiple starts than running constantly.

Tanque

I would imagine that it isn't simple system to model in order to determine power requirements.

parameters I could see contributing to the power requirements

weight/ mass ( a vector analysis comes to mind with regards to forward/ rearward motion)
coefficient of friction between the tracks and surroundings
heat lost transmitting power from the source to the tracks/ power consumption of all the components of the running gear

Not certain regarding the dynamics or how the calculation would need to be to allow for the dynamics of the unsprung mass of the road wheels and track.

There nay be models already designed for full size machines to encompass your question.

You could probably do an approximate calculation by determining how many watts of power are consumed per minute of operation ( or joules= watts/second )
by multiplying volts x amps drawn under load conditions. By interpolating over the period of time you want to operate you should be able to
determine fairly well what capacity of battery you require.

That's one of the beauties of gas/ glow models I just fuel 'er up and go...of course there's still the absolute amount of power required.
It doesn't really matter what the source is. My typical Webra .40 in good condition is rated at about .7 hp at top speed which is about 529 watts ( 756 watts/ hp x.75)

Jerry

lposter

What started me on this was the treatment given by Clark and Owings in "BUilding Robot Drivetrains" - they have a fairly standard analysis of the whole thing for determining the size of motor needed to power robots under various conditions (uphill etc etc).

BUt they are focussed on wheeled robots.

Ihad no problem with their treatment of the matter until I realised that a wheeled robot "weighs" exactly what the bathroom scales says it does and that maybe is not true for a tracked vehicle.

People might say this has no bearing on anything but a plastic tank weighing a kilo without tracks may weigh 2.5 kilos with metal tracks. If half the weight of the metal tracks is not contributing to the downward force exerted by the tank - then its a significant portion of the entire assumed mass of the tank which enters the calculation.

In addition, once the required torque to get the thing moving is determined (again relaint in th emain on an accurate measure of the mass of the tank), then the power needed can be determined....

Once the power is gotten (for a given voltage) the parameters like maximum current required and battery capacity needed can be calculated.

BUt again...all this was for wheeled robots not tracked.

Anyone know of any books where a full treatment of the above problem is given for tracked vehicles? because deriving it for myself is beyond my capacity.

Could be an interesting problem for some high school physicist with an interest in tanks.....

p

Tanque

I remember a physics question that really was deceptive simple in statement.

A cubic block of ice sits on the top of a hemispherical dome.

It begins to slide off.

Assume no friction between ice block and dome.

At what point does the ice block leave contact with the dome...

Jerry

kubel-head

Iposter,

heavyaslead

A rather simple battery - motor capacity measurement would be to set the motors to stall and draw the battery till it expires.

Weight goes to infinity when under stall conditions, the worst case.

That would determine max loading and time to run, just pick a battery size to maximize time.

lposter

Im not much of a whizz meself......but I just used the weight of "half" the track to illustrate my point. I appreciate as you correctly state that it is unlikely to be half at all.
This is true......but the idea of calculations is to determine motor sizes etc before purchase.....I tried today with a kitchen scale.....measuring the weight of the tank off the ground but with the tracks over teh sides of the scales and then slowly lowering the scales until th etracks were on the ground but not the road wheels.Didnt go very well at all for the lack of a controlled way of raing and lowering the scale in mm amounts.....p

Perry S.

Ths actually depends on the types of calculations. It takes the same amount of energy to move a tanks mass regardless of it is sprung or unsprung weight. Rememberr your high school physics, kinetic energy. To raise the tank up an incline it takes the same amount of energy regardless of sprung or unsprung weight. Remember potential energy. So depending on the types of formula in the calculation you may be alright using the total mass of the tank.

I think what you are realy trying to do though is to determne the losses in the system because the rest is handled by the above. Primary in my opinion is the work required to bend the tracks about their pins. That is why thinner tracks have less losses than wide tracks, there are less interfaces where the pins rub against the links. I do not think ground resistance matters here, you want it high, that is your traction.

I have not read the book you cited but if you could post the formulas it seems like there are enough smart people in this thread to answer your question difintevely. Judging by the responses I have seen so far.

Chances are;

PE = m*g*h
KE = 1/2*m*V2
F = m*a

Will get us most of the way there.

Perry

lposter

Itried to write this out in word as the editor treats me badly these days.


Iam avoiding symbols simply due to me not trusting the editor.






MY basic question is to the validity of Equation 4 with respect to tracked vehicles.

AS other posters correctly pointed out the forces acting on the tank (both in relation to tank-track forces. tank - ground forces, track-ground forces, track - drive wheel forces, track-track forces) probably cannot be adequately represented by models (which the above equations are)representing a simpler system.

Hence my original question as to whether the weight of a tank (ie. mass x acceleration due to gravity) is as simple as just weighing it.

Does anyone know of a similar treatment fro tracked vehicles?

Please dont tell me its irrelevant and just shove in the biggest motors you can find........

p

heavyaslead

Moving mass is different from static mass and getting static mass to move (accelerate) is still another parameter.

I'm getting confused. What exactly are you trying to do?
Size a battery?
A motor?
Or achieve some power to weight ratio?

Or are you just going through the math exercise

lposter

NOpe...Ive got motors and I know the torque.
Now I want to know what size tank they can move, inclines etc etc.
So I need the formulae....
p