I called on one of the subcontractors to size a geared motor. The problem is that I am not offered the same thing at all. If the output speed and power are close in both cases. One offers me a torque of 280 Nm and the other with a torque 3 times higher! I am surprised to have such a big difference between my two suppliers.
I would therefore like to determine the characteristics of the geared motor myself to have a more critical look at what is offered to me. I am calling on you to know the method to follow!
I have a load of 8t to be transferred on a flat ground at a speed of 2m/min. The drive is done by a pulley-rack system. Gear Feature: Module 8, Number of Teeth 15.
I think that an important element is missing, which is friction.
8 tons on a wheeled chassis is not the same as moving 8 tons on concrete.
Is it on rails / guides for example?
What is important is to know how much effort you need, for speed, if your motor is managed by a variator, it can be adjusted at worst as long as it has enough torque.
After of course who can do more can do less, but your engine must be able to stop the load in motion too.
I should have said it, it's true. The load is on wheels guided by a rail effectively. Seen with the suppliers I will install a variator to avoid brutal efforts and the brake is provided.
Precisely for me, the effort required to move the load corresponds to the output torque??
PS: Having contacted my supplier who offers me the lowest torque telling him that I had another proposal with a torque 3 times higher, he explained to me that having a torque that was too high was not good either and that it could break the reducer. And that in order to have something reliable, you had to pay attention to the "use factor", which must be close to 1, a little higher.
First of all, the big mistake that I see hundreds of times on another forum is to confuse input and output power.
The golden rule is that input power = output power - (friction losses).
What you need to understand is that the reduction motor, in a way, allows you to shorten the time , which means that instead of having your power in Joule in an hour or a minute, you have it in 10 times less time (if you have a 10/1 reduction gear).
You also have to take into account the N/m number of your sprocket.
But your problem is not there because only the inertia to overcome at starting and braking is to be taken into account, (once the mass is in motion it's almost nothing to move your mass) Inertia at start for your engine and braking for the teeth of the pinion and your rack if the engine is stopped instantaneously, i.e. without electronically controlled slowdown.
You don't indicate the speed of movement, which is crucial for inertia because a mass m (kg) that we accelerate with a constant acceleration a (m/s/s) requires an effort: F(N) = m. A until the desired speed is reached. The effort to overcome braking inertia is equal to that of acceleration. So time and speed are elements to be taken into account outside of the mass.
You also need to indicate how many rollers support your load and whether it slides on rails (or the ground) horizontally or vertically.
Kind regards
EDIT: the brake is not necessary because with a geared motor (example screw reducer) there is an irreversibility depending on the number of stages. The brake is only useful if your mass has to be stopped and especially kept on a slope. The choice of brake may unnecessarily increase the cost of the package. The start-up and braking assistance can also be superfluous depending on the speed and the reduction ratio.
Do you have a time constraint to make the move or to reach the speed of 2m/min, this will determine your acceleration/deceleration and therefore the max torque if I'm not mistaken (distant memory for me ;) )
friction coef = b/R = 2/32.5 = 0.06. I take 0.07 to be safe
Ff = 8000*9.81 * 0.07 = 5588.8 N
So my strong total Ft = F + Ff = 5641.6 N
From there I calculate the torque necessary to set the load in motion with a gear wheel m=8 and Z=15:
C = Ft + r = 5641.6 + 0.06 = 338.5 Nm
w=v/r = 0.55 rad/s and P = C.w = 186.2 W, on these two data I am in agreement with the suppliers.
On the other hand, the first geared motor has a torque of 280 Nm at the output. He didn't have any information about the acceleration, maybe that explains the quarter?
The other supplier gives 900 Nm of torque. Would the difference seem abnormally high to me?
Anyway, all that, assuming that I have done the right method and the right calculations!