Traction Confusion: Overcome Wheel Spin & Improve Your Grip

[Train Man]

TL;DR

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?

 

[jack action]

The first number, $4312\ N.m$ (and not $4312\ N/m$), represents the torque needed from the locomotive engine such that you can produce $9500\ N$ of driving force at the wheels.

The second number, $16\ N.m$, represents the torque needed from the locomotive engine such that you can produce $285\ N$ of driving force at the wheels. ($0.015 \times 9500 N \times 0.20 = 285\ N$)

The problem in these equations is that $TTE$ in the first part represents a driving force pushing the locomotive (horizontal force), and the $TTE$ in the second part represents the normal weight of the locomotive (vertical force). It is confusing that they share the same name (and it is not a 'power', it is a force in both cases) while representing two different things.

 

[Train Man]

So thus in the second part the wheel radius is not taken into account. ($0.015 \times 9500 N \times 0.20 = 285\ N$) Might that be correct?

 

[jack action]

It is if you want to find the wheel torque required. Torque $T$ is related to the wheel radius $r$ through:
$$T= Fr$$
In your first equation, $F = R_f \times TTE$;
In your second equation, $F = \mu \times TTE \times f$.

But, as I said earlier, $TTE$ doesn't represent the same thing in both equations. It's a very confusing notation.

 

[Train Man]

Thanks jack actio, I will change the TTE in the second equation to reflect the vehicle (GVW) weight.

The torque requirements of the combination equates to 10,792 N.m at the road surface as the image outlines. This includes for an operational speed of 25km when fully loaded and a 5 degree slope as accounted for in the earlier calculations.

Firstly, is this reflective of what the 'Peak Torque' (PT) requirement would be for the combination? I might then divide this figure by the gear ratio to find the 'Peak Torque' requirement for the drive motor. Eg 10,792 divided by 31 = 348 N.m. Then that might indicate that a motor with a 380 N.m PT performance would be ok.

Secondly, the term 'Continuous Torque' might be that which is reflective of traveling on flat ground when loaded. The combined 'Rolling Resistance' figure 28,500 N contains elements for RR + GR and AF. How much of RR + GR and AF get taken into account in establishing what the 'Continuous Torque' might be?

 

[Lnewqban]

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?

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?

 

[jack action]

The first equation says that you want the locomotive to produce $9500\ N$ (It seems to be a design requirement). This is the force that will go against $RR + GR + AF$ (assuming rolling resistance, gravity resistance, and aerodynamics force). If the driving force is larger, then the locomotive will accelerate ($TTE - (RR + GR + AF) = ma$).

Multiplying that value with the wheel radius gives you the required wheel torque.

To get the required motor torque to provide the desired force ($9500\ N$), you need to divide the wheel torque by the drivetrain efficiency ($0.85$). Your equation multiplies it which doesn't make any sense. (Unless I don't understand clearly what it represents.)

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 $3 \times 9500 = 28500$).

How much of RR + GR and AF get taken into account in establishing what the 'Continuous Torque' might be?

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]

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).

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.

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. However it will only be the loco that provides all of the pulling ability.

Correction requirement you drew attention to as follows:

 

[Train Man]

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?

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.

 

[jack action]

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.

OK.

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.

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.

However it will only be the loco that provides all of the pulling ability.

Then only the locomotive wheel radius matters. Assuming $28500\ N$ of resistance for the train, then the motor torque required is $28500\ N \times 0.534\ m / 0.85 = 17905\ N.m$

I got to thinking about converting the loco engine (diesel burner) to electric by removing the engine and installing an electric motor.

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]

You need to find the aerodynamics force for the whole train, not each wagon.

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.

 

[Train Man]

Matching motor speed is then just a matter of selecting the proper final gear ratio and the torque outputs will automatically match.

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.

So thus I can only compute an average gear ratio's across what exists between (a), (b) and (c). The manual does give me the distance traveled in each gear for the large agri tyre at max recommended engine revs of 2200 RPM. Taking the distance traveled from the manual and computing with rolling circumferance and rotations I establish a average gear ratios.

I then apply the ratio info to the loco tyre as fitted to establish a new set of km/h speedsfor the train combination. So thus I think your 'final drive' mention is the same as what I call my average gear ratio.

So if the electric motor output can be managed to reflect the recommended max 2200 RPM of the old engine then we are sucking diesel!

 

[jack action]

Say your diesel train puts out 230 kW @ 1500 rpm when riding at 25 km/h (6.9444 m/s).

You also know that the wheel RPM at 25 km/h is:
$$\frac{6.9444\ m/s}{0.534\ m} = 13 \frac{rad}{s} \left[\times \frac{60 \frac{s}{min}}{2\pi \frac{rad}{rev}} = 124\ rpm \right]$$

The overall gear ratio between the engine and the wheel must therefore be 1500 / 124 = 12.1:1. (Although it is not important to know that)

The power at the wheel is conserved (minus the efficiency losses), thus the wheel power is 230 kW * 0.85 = 195.5 kW. (Although it is not important to know that)

The wheel torque must then be 195 500 W / 13 rad/s = 15 038 N.m. (Although it is not important to know that)

Now say you replace your diesel engine with an electric motor that produces 230 kW @ 3000 rpm.

The wheel RPM stays the same at 25 km/h, thus the new overall gear ratio between the electric motor and the wheel must be 3000 / 124 = 24.2:1.

The wheel power is still the same at 195.5 kW (maybe slightly different if the new gear ratio set up has a different efficiency compared to the old one).

The wheel power and RPM being the same, the wheel torque is also the same (15 038 N.m), so from the point of view of your wheel, everything is the same. If it was enough for your diesel engine, it will be enough for your electric motor.

 

[Train Man]

Say your diese

The tractor in its old agri set up at 99HP/75kW provided for a max torque of 360Nm. I assume that is provided at the 2200 RPM point and nothing has changed.

Does the following read correctly...?

Now substituting into the 'jack action' text - diesel the train puts out 75 kW @ 2400 rpm when riding at 25 km/h (6.9444 m/s).
You also know that the wheel RPM at 25 km/h is:

thus ...the 6.944m/s devided by the 0,534m as the jack action formula gives 124 rpm

The overall gear ratio between the engine and the wheel must therefore be 2400 / 124 = 19.3 : 1.

The power at the wheel is conserved (minus the efficiency losses), thus the wheel power is 75 kW * 0.85 = 63.75 kW.

The wheel torque must then be 63 750 W / 13 rad/s = 4904 N.m.

Now say you replace your diesel engine with an electric motor that produces 75 kW @ 3000 rpm.

The wheel RPM stays the same at 25 km/h, thus the new overall gear ratio between the electric motor and the wheel must be 3000 / 124 = 24.2:1.

The wheel power is still the same at 63.75 kW (maybe slightly different if the new gear ratio set up has a different efficiency compared to the old one).

The wheel power and RPM being the same, the wheel torque is also the same (4904 N.m), so from the point of view of your wheel, everything is the same. If it was enough for your diesel engine, it will be enough for your electric motor.

The top end gears for 1st, 2nd, 3rd and 4th seem to be 17, 24, 31 and 49:1 as per my gear ratio calculations.

 

[jack action]

Does the following read correctly...?

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.