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Train Man
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- I am following some resources from the net to work out Drive Wheel Torque at the wheelend/road surface but getting two totally different answers. As image attached. Is either of them correct?
Those posted resources from the net are a mess!Train Man said:Summary:: I am following some resources from the net to work out Drive Wheel Torque at the wheelend/road surface but getting two totally different answers. As image attached. Is either of them correct?
All of it. If you have a total resistance of ##28500\ N##, your locomotive driving wheels must produce that much force to maintain a steady speed.Train Man said:How much of RR + GR and AF get taken into account in establishing what the 'Continuous Torque' might be?
For clarity the loco weight = approx 3500kg's. The coaches when loaded are = 3500kg's each. So the total kg's of the combination when loaded is 10.500kg's.jack action said:The combination torque you presented in your last post seems to require that the 'loco' and the two 'coaches' all require to produce 9500 N each. I'm not sure this is what you are looking for (but I think it is, since 3×9500=28500).
They may be a mess but I'm capable of further adding to a mess as I'm not really a Newton head. After the Environmental Summit in Scotland I got to thinking about converting the loco engine (diesel burner) to electric by removing the engine and installing an electric motor. Physically it seem possible but I was trying to establish some parameters by looking at numbers on paper so to speak to establish the size of motor and battery pack required. The secret video at this link better explains. However with this Covid thing nothing can be progressed or planned as we have been sitting idle since it commenced.Lnewqban said:Those posted resources from the net are a mess!
What are parameters of your specific problem, and what are you trying to calculate?
How many driver wheels?
OK.Train Man said:For clarity the loco weight = approx 3500kg's. The coaches when loaded are = 3500kg's each. So the total kg's of the combination when loaded is 10.500kg's.
Although that might work for RR+GR, it wouldn't work for AF. You need to find the aerodynamics force for the whole train, not each wagon.Train Man said:I worked out the RR + GR + AF for the loco and found it to be 9500 N and then applied it to the 3 vehicles that make up the combination hence the 28500 N.
Then only the locomotive wheel radius matters. Assuming ##28500\ N## of resistance for the train, then the motor torque required is ##28500\ N \times 0.534\ m / 0.85 = 17905\ N.m##Train Man said:However it will only be the loco that provides all of the pulling ability.
Then you should look for replacing the diesel engine with an electric motor that can deliver a similar power output (kilowatts are units equivalent to horsepower). Matching motor speed is then just a matter of selecting the proper final gear ratio and the torque outputs will automatically match.Train Man said:I got to thinking about converting the loco engine (diesel burner) to electric by removing the engine and installing an electric motor.
Because this train combination has a max operating speed of 25km/hr on city streets I made no allowance for aerodynamic drag in the calculations. Again I read somewhere on the net that drag does not come into effect until higher speeds are involved. Might that be an incorrect assumption.jack action said:You need to find the aerodynamics force for the whole train, not each wagon.
I do not know the gear ratios of the tractor. Only that it has 4 gears (1st, 2nd, third and forth) with a high, medium and low range. So lying between the input shaft and the road surface there is in existence (a) a standard gear box, (b) a high, mid and low range box and the (c) differential gear on the rear axle.jack action said:Matching motor speed is then just a matter of selecting the proper final gear ratio and the torque outputs will automatically match.
jack action said:Say your diese
Yes, it does. But is the diesel engine really at 2400 rpm when riding at 25 km/h and does it really output 75 kW (i.e. maximum fuel delivery)? It is rarely the case when at "operational" speed to be at maximum output and maximum rpm, but it could be possible in your case.Train Man said:Does the following read correctly...?
Traction confusion refers to the phenomenon of wheel spin, where a vehicle's wheels lose grip and spin on the surface, resulting in a loss of control and movement. This can be caused by various factors such as weather conditions, road surface, and vehicle weight.
Wheel spin can significantly impact a vehicle's performance by reducing its ability to accelerate, steer, and brake effectively. It can also cause excessive wear and tear on the tires and other components of the vehicle, leading to potential safety hazards and increased maintenance costs.
One way to overcome wheel spin is by using traction control systems, which use sensors to detect wheel spin and adjust power delivery to the wheels accordingly. Other methods include using winter tires for better grip in snowy or icy conditions, reducing vehicle weight, and avoiding sudden acceleration or braking.
Yes, wheel spin can be dangerous, especially in situations where a driver loses control of the vehicle. It can also increase the risk of accidents, especially in wet or slippery road conditions. Therefore, it is essential to take necessary precautions and address any traction confusion issues to ensure safe driving.
Improving your vehicle's grip can be achieved by regularly maintaining your tires, ensuring proper tire pressure, and using appropriate tires for different weather conditions. It is also essential to drive cautiously and avoid aggressive maneuvers that can cause wheel spin. Additionally, using traction control systems and other technologies can also help improve grip and control on the road.