I am trying to find the number of tyres for the given data:
Vehicle GVW = 7500 Kg
Coefficient of friction = 0.72
Torque given to the wheels = 4073 N-m
Probable tyre selection = 7.5R16
Tyre can support a normal load of = 1315 Kg
I am unable to get realistic results. Any help will be appreciated
Yes, I need traction on the tyres and whether the tyres will be able to provide them for the given amount of torque.
Hence, the number of tyres calculation.
Tyre 7.5R16 has a diameter of about 600mm (equals to a radius of 0.3 m).
max. friction force at one tyre is: 13150 x 0.72 = 9468 N
max. torque at one tyre is: 9468 x 0.3 = 2850 Nm
==> That means, that 2 tyres should be sufficient to transfer a torque of 4073 Nm.
Kind regards,
Mario
One clarification here is that 4073 N-m of torque is required to be transmitted by each wheel if 8 driven wheels are considered which is impractical, though.
Actual total torque requirement is of 32590 N-m, which sums up from Starting resistance force + Acceleration force + Gradient resistance (25%).
Also depending on the number of wheels, it will have a normal reaction, in turn giving frictional force to transmit torque. So 13150 N cannot be taken.
Can you build your help on this.
You have given a total vehicle mass of 7500 kg equals to 75000 N, and a friction factor of 0.72. In this way, the maximum friction force is 75000 x 0.72 = 54000 N (This value is independent from the number of wheels.)
Considering a tyre radius of 0.3 m, the total maximum torque provided to all tyres is 54000 x 0.3 = 16200 Nm. Result: With these boundary conditions, it is not possible to transfer a higher wheel torque, than in sum 16200 Nm. If you want to transfer a total torque of 32590 Nm, you have to increase the friction factor, or the mass of the vehicle, or you can change the wheel diameter.
Kind regards,
Mario
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