Torque and "torque slew rate"

In a recent thread, some discussion about torque and "torque slew rate" came up. I thought the discussion deserved a separate thread. This may be all wrong, but I'll go first:

So what is “torque” anyway?

Torque is a rotational force; a force which (in sim racing) is applied perpendicular to the radius of a steering wheel. It’s what your wheelbase applies to turn the wheel or resist your hands as you are also applying turning force (also torque) to the wheel.

So what’s a Newton-meter (Nm), the unit of torque? Let’s start with “what’s a Newton?”

A Newton (N) is a unit of force. It’s the amount of force required to accelerate one kg of mass at the rate of 1 meter per second squared.

Because the time (seconds) is squared, we are talking about acceleration, not velocity.

Now let’s take a look at how torque, denoted as tau, with units of Nm (Newtons*meters), is applied in the sim racing world. We need to know three things: The wheel radius, its mass ("weight") at its radius, and the torque applied to it.

(1) Let’s say your wheel is 30 cm in diameter (quite typical). So its radius is 0.15 m.

(2) Let’s say for a typical wheel and button box combo, we call the lumped mass at the rim's radius to equal 0.5 kg.

(3) Let’s say you have a wheelbase set to apply 10 Nm torque.

How much force is applied at the rim? Equivalently, how much force must you apply to prevent the wheel from turning? Assuming that the torque is always applied perpendicular to the radius of the rim (as is normal in sim racing):

Torque (tau) = radius (r) * force (F)

Or:

(10Nm)=(0.15m)*F

We want to solve for the force F, so:

F=(10Nm)/(0.15m)=66.67N

For those accustomed to English units, one Newton equals about 0.225 pounds (of force, not weight!), so that’s about 15 lbs of force spinning madly while you wrestle it.

Given this force, how fast will the wheel be turning, starting from a stand-still, after one second?

F=mA

So, solving for the acceleration A:

A=(66.67 kg*m/s^2)/(0.5 kg)=(133 m/s^2)=(436 ft/s^2)

If s=1 (one second), then that’s 436 ft/s velocity, which equals 297 MPH!!!

Well, OK, that’s probably not going to happen, because the motor has some maximum speed of operation that is likely to be considerably lower (e.g., the electrical impulses that activates the motor are only so frequent at their peak, and the motor won’t go faster than that).

One take-away from the above is that the maximum torque (when freely spinning unopposed) can only be applied for a very short time (likely less than one second) before the system has no choice but to reduce the applied torque to close to zero, i.e. once maximum RPM is reached. But needless to say, the wheel can turn very, very quickly and with a lot of force applied at the small radius of a typical wheel. And that’s with “only” 10 Nm applied torque. No wonder people can hurt themselves with these wheels!

Now let’s look at “torque slew rate.” This is a figure of merit supplied by at least one manufacturer (SimuCube). This is presumably the maximum rate at which the applied torque can change. Their three systems are noted as having:

Sport: 17 Nm max, 4.8 Nm/ms slew rate
Pro: 25 Nm max, 8 Nm/ms slew rate
Ultimate: 32 Nm max, 9.5 Nm/ms slew rate

Unless artificially limited in firmware, one might expect slew rate to be directly related to max Nm. And indeed this is roughly true, e.g. how long does each take to reach maximum torque from a standstill?

Sport: 17/4.8=3.5ms
Pro: 25/8=3.1ms
Ultimate: 32/9.5=3.4ms

So all three can theoretically reach their maximum torque in about 3.3 ms. A rapid maximum-force tank-slapper (+max to –max torque in a few ms) would likely be very difficult to control.

In any event, the actual achievable torque slew rate is dependent on the limits of the motor itself, the wheel-base’s firmware limits, any in the OS, any particular game’s feedback limits, and the weight and diameter of the wheel and button box, whose moment of inertia opposes acceleration.

So another take-away is that, for maximum sensitivity in force-feedback, you should prefer a light-weight and small-diameter wheel.

I hope this is a helpful starting point. It's semi-likely that I've made an egregious mistake somewhere, so please point out any errors (gently). I reserve the right to fix errors in this post. Thank you.

Updated to fix various errors thanks to Neilski (hopefully without introducing too many new ones!). Thanks!
 
Last edited:
one Newton equals about 4.45 pounds (of force, not weight!)
You've accidentally got the ratio inverted, for a net error of a factor of 20 in the number of lbs force ;)
A=(66.67 kg*m/s^2)/(0.15 kg)
You've accidentally used the 0.15 figure again here instead of 0.5 kg which was the stated mass of the wheel. It's still gonna end up spinning pretty fast though.
By two seconds, because time (s) is squared, it will be going 4 times faster
Nope, velocity is linear with time. It's distance travelled (or number of revolutions of the wheel) that is quadratic with time.
This is presumably the maximum rate at which the applied torque can change.
Yup, and it's essentially linked to the rate at which you can change the current in the motor, which itself is linked to how much voltage is available (and the inductance of the motor winding) combined with how close to that limit the drive electronics can push it.
So all three can theoretically reach their maximum torque in about 3.3 ms.
Yup. For fun, you might want to work what angle the wheel turns through if (starting from rest) you ramp it to max torque, back down to zero, immediately to negative max torque, and back to zero again, all at max torque slew rate. It should finish up at rest again in a new position. I don't think it'll have turned very far but I haven't made the calc myself.
 
Upvote 0
Thanks Neilski!

I think I'll leave that last calculation for others such as yourself, since I think some calculus is needed to do it properly, and I'm too rusty to rush into that. It's an interesting problem, though, and it may be enlightening.
 
Upvote 0
I think I'll leave that last calculation for others such as yourself, since I think some calculus is needed to do it properly, and I'm too rusty to rush into that. It's an interesting problem, though, and it may be enlightening.
Well, I'm rusty as hell at calculus myself, but it turns out that symmetry allows you to work it out without any actual integration. (Btw, I would normally do a calc like this using moments of inertia and in terms of angle, but I've left it in terms of linear velocity and distance to match your calcs above.)

If you choose the 25 Nm motor with 8 Nm/ms torque slew rate, with the same 0.15 m radius wheel of 0.5 kg mass, it goes like this:

F=torque/radius = 25/0.15 (i.e. force at rim)
If we ramp from 0 to F and back to 0 in minimum poss time, the average force during that interval is clearly F/2. (This is just half of the total story, remember - the second half of the story takes us from 0 to -F and back to 0.)
The time for that up and down pair of accel ramps is just t=(2*25/8e3) seconds
Thus final velocity is just (using F=m*a and v=a*t): a*t=(25/0.15/2/0.5)*(2*25/8e3) = 1.04 m/s
Then, by symmetry, the average velocity during the first pair of up/down ramps is just half of the velocity at the end of the down ramp (i.e. the end of the acceleration phase); then comes the deceleration phase, which lasts the same amount of time and brings the wheel to rest, so the distance moved is just twice as much.

Twice the distance moved during the acceleration phase is therefore (I've deliberately left in the cancelling division and multiplication by 2):
2* (25/0.15/2/0.5)/2 *(2*25/8e3)^2 = 6.5e-3 (6.5 mm).
So yeah, the wheel rim moves a surprisingly small amount.

Also, since the slew rate is on the bottom of the equation and is squared, you can instantly see that if you halve the slew rate, the wheel will travel four times as far during these max torque excursions.

To be paranoid, I checked it with my horribly rusty integration skills, made dozens of dumb errors along the way, and finally proved to myself that the symmetry assumptions were justified. I also learned the (oh-so-useful :D) fact that the wheel travels 5 times as far during the down-ramp of the torque as during the up-ramp. (That's about the only useful extra info I can see that you get from doing the brute-force integration.)

NB: it's highly likely that I botched at least part of this ;) :roflmao:
 
Upvote 0
Awesome!

I was thinking that calculus might not be necessary after all too (the F/2 bit) - or that it would be a trivial case to integrate anyway. I was making a bit of progress, but then girlfriend-trouble caused a serious interruption to my thoughts. :laugh:

Anyway, I hadn't anticipated the down-ramp being so different, so that's interesting!

Fun! Thanks!
 
Last edited:
Upvote 0
TL; DR:
( in fact didn´t fully understand)

if you blow it and suffer a "tankslapper", get your hands off the wheel NOW!!

I learnt that very quickly with the SC2.

And please, keep on nerding ;)

MFG Carsten
 
Upvote 0
Regarding "tank-slappers": Yes, I've learned that the hard way too. Ow!

It's early days for me, but I've noticed quite some differences in the levels and characteristics of danger that different games cause.

"Wreckfest" doesn't mess around - it throws a lot of power at you (as it should with a game like that!), but thankfully it rarely totally jerks the wheel. It's mostly strong low-frequency effects. That way, you can turn up the power and go total cave-man on it without much concern about your thumbs.

By contrast, "Automobilista 2" can be extremely nasty! It's not so much the driving or the crashing. It's the post-crash. It gets confused and wants to recenter or something (often repeatedly), using maximum force and speed, when all you want to do is gently drive off from a stop. This most frequently happens when you wind up alongside a fence or other interference. It's a bug, in my opinion, and it's so violent that I've really had to turn down the power just to avoid these post-crash scenarios. (In most other ways, it's my current favorite sim.)
 
Last edited:
Upvote 0
Thanks for the info, I understood some of it! Please correct me if I'm wrong in terms of sports comparison to pro. Does this mean that both motors are approximately the same speed in terms of slew, just one "jerks" harder? IE the sport isn't really any slower, it's just not as aggressive. Was having this conversation on reddit today.
 
Upvote 0
It means that the "torque slew rate" is about proportional to the motor's torque rating. Go up from the sport to the pro, or pro to the ultimate, and both scale together. Double your pleasure and fun.

Power is easy to feel and compare, but what torque slew rate means in game is a more subtle question. The next steps might consider motor speed ratings, update rates, game effects, filters, etc.

I think most people would be very happy with any of them.
 
Last edited:
Upvote 0

What are you racing on?

  • Racing rig

    Votes: 528 35.2%
  • Motion rig

    Votes: 43 2.9%
  • Pull-out-rig

    Votes: 54 3.6%
  • Wheel stand

    Votes: 191 12.7%
  • My desktop

    Votes: 618 41.2%
  • Something else

    Votes: 66 4.4%
Back
Top