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amount of soil slip (slip percentage)
- Thread starter yabande
- Start date
Surely an equation for the braking action can be written on the time variable
v(T)=v0-alfa·T if decceleration can be assumed constant
if not constant
V(T)=v0-integral between 0 and T of alfa(t)·dt
then
s(T)=integral between 0 and T of v(t)·dt
Now we only need to state a formulation for the decceleration and problem solved. Let's think a bit how except someone comes handy with the solution...
v(T)=v0-alfa·T if decceleration can be assumed constant
if not constant
V(T)=v0-integral between 0 and T of alfa(t)·dt
then
s(T)=integral between 0 and T of v(t)·dt
Now we only need to state a formulation for the decceleration and problem solved. Let's think a bit how except someone comes handy with the solution...
Got it?
From impulse equation a force F standing dt causes a decrease in velocity dv
equation
F·dt=m.dv
ir if you want a velocity of the amount of dv is cancelled by the application of a force F during time dt.
hence
dv/dt=decceleration=alfa=F/m
when the force is a friction force
F=mu·N=mu·m·g
hence
alfa=F/m=mu·g
hence decceleration is the friction factor times the gravity acceleration.
Substitute above and finished, except I am in error.
From impulse equation a force F standing dt causes a decrease in velocity dv
equation
F·dt=m.dv
ir if you want a velocity of the amount of dv is cancelled by the application of a force F during time dt.
hence
dv/dt=decceleration=alfa=F/m
when the force is a friction force
F=mu·N=mu·m·g
hence
alfa=F/m=mu·g
hence decceleration is the friction factor times the gravity acceleration.
Substitute above and finished, except I am in error.
- Thread starter
- #4
I need an amount of slip which is function of tractive force.
tractive force is a force that the terrain can support before slip happens.
Slip equals to s= 1-(actual speed of tire/theorical speed of tire). i have got the theorical speed but is very difficult to calculate the actual speed. Therefore must be another solution for it.
Forget about breaking or acceleration. Just imagine a vehicle going up the sandy slope with constant velocity. How much is slip? Should be some how related to slope angle and weight of vehicle or net tractive force.
It can advance on inertia whilst the rotational speed of the wheels is such that entirely overcomes the angle of inner friction in the sand, at with situation, the vehicle wouldn't be impulsed anymore ahead by the wheels. Should you impart instantaneously such speed to the wheels whilst the vehicle stays, sand would be being expelled without any advance of the vehicle.
Only for the intermediate situations between full splippage and no slippage your question seems to have some numerical interest. In these situations you have a dynamic friction factor but the wheel still has not such speed (or rotational energy) that the inner friction at the soil have been overcome. In this situation you have a dynamic friction force yet through this dynamic friction the wehicle can still get reaction to the torque imparted to the wheel to produce further advance, to the limit of the dynamic friction becoming equal to the angle of inner friction, moment at which the vehicle will loss its archimedian point of support for advance.
Starting from this we can think of a number of models that give the slippage. Will try.
Only for the intermediate situations between full splippage and no slippage your question seems to have some numerical interest. In these situations you have a dynamic friction factor but the wheel still has not such speed (or rotational energy) that the inner friction at the soil have been overcome. In this situation you have a dynamic friction force yet through this dynamic friction the wehicle can still get reaction to the torque imparted to the wheel to produce further advance, to the limit of the dynamic friction becoming equal to the angle of inner friction, moment at which the vehicle will loss its archimedian point of support for advance.
Starting from this we can think of a number of models that give the slippage. Will try.
Hate to say it but the Agriculture guys might be the best to ask . . . they are into big tires and earth for their equipment. I've been keenly surprised by what some of their studies are about - i.e., using a double rod nuclear density meter for field densities . . .
The total net torque applied to the wheels when not slipping produces a reaction force on the soil that is employed in overcoming resistance to advance and produce acceleration if over that.
But in a situation in which slippage is occurring (the intermediate situation of interest above referred), part of the energy is employed in the dynamic friction, or if you want, what would be the normally applied load, F=T/R, where T net torque applied, R radius of the wheel, F force at the interface, F, now, has a loss, again mu·N=mu·m·g.
Now we need to define parameters that quote the problem. For a complete chart, our vehicle running would start running at so low acceleration that produces no slippage at all (0 slippage), make grow (accelerate) its speed to reach one that attains some (constant?) dynamic friction (0 to 100 slippage), to accelerate in end at such high rate that even the angle of inner friction is overcome and slippage is 100%.
Whta can make the wheel slip is excessive force respect the friction at the interface; a vehicle at constant speed whatever its speed will be thought not to have slippage at all. This excessive force hence can be obtained from at rest, or from any ongoing speed, applying enough acceleration. So we would have:
T/R=F force from the net torque (after inner losses) must equal to sum of
wind reaction at the speed
force from the friction loss from ongoing slippage
force available to produce net acceleration, for a case where the vehicle still is able to accelerate in spite of the slippage,
or if you want, you would be getting for the case as net force to produce the wanted effects (overcome wind and produce acceleration) that of resulting from net torque less than employed in overcoming dynamic friction, once slippage is present.
the term for friction loss from ongoing slippage wouldn't be present in the case where the accelerations do not cause slippage.
The wheel may still rotate at the same speed that when no slippage was present, but the torque is not employed in the same way. Advance (speed) is diminished in the same proportion than dynamic friction forces mu·m·g represent respect F=T/R, T being net torque.
But in a situation in which slippage is occurring (the intermediate situation of interest above referred), part of the energy is employed in the dynamic friction, or if you want, what would be the normally applied load, F=T/R, where T net torque applied, R radius of the wheel, F force at the interface, F, now, has a loss, again mu·N=mu·m·g.
Now we need to define parameters that quote the problem. For a complete chart, our vehicle running would start running at so low acceleration that produces no slippage at all (0 slippage), make grow (accelerate) its speed to reach one that attains some (constant?) dynamic friction (0 to 100 slippage), to accelerate in end at such high rate that even the angle of inner friction is overcome and slippage is 100%.
Whta can make the wheel slip is excessive force respect the friction at the interface; a vehicle at constant speed whatever its speed will be thought not to have slippage at all. This excessive force hence can be obtained from at rest, or from any ongoing speed, applying enough acceleration. So we would have:
T/R=F force from the net torque (after inner losses) must equal to sum of
wind reaction at the speed
force from the friction loss from ongoing slippage
force available to produce net acceleration, for a case where the vehicle still is able to accelerate in spite of the slippage,
or if you want, you would be getting for the case as net force to produce the wanted effects (overcome wind and produce acceleration) that of resulting from net torque less than employed in overcoming dynamic friction, once slippage is present.
the term for friction loss from ongoing slippage wouldn't be present in the case where the accelerations do not cause slippage.
The wheel may still rotate at the same speed that when no slippage was present, but the torque is not employed in the same way. Advance (speed) is diminished in the same proportion than dynamic friction forces mu·m·g represent respect F=T/R, T being net torque.
I think the last paragraph is in error. Speed is diminished in what the reduced available force to produce wanted effects (overcome wind, accelerate) means to speed. A reduced traction force with the same (previous) wind opposition will means instantaneously more loss of speed than that the proportional to the instantaneous loss of available reaction (equalling the reduced force after friction) available for useful effects.
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