Torque sensor question

bluecat

Well-Known Member
The formula is Power (In Watts ) = Torque (in nM) x Angular speed(In RPM) /9.55 . This is same formula above, we just converted it to be used with rpm, 2 PI x RPS = (2 PI /60) x RPM = RPM/9.55.
Plug the values in you get,
8x300/9.55 =251.3 ~ 250 W.

I fully agree with this (the formula shows the dependencies of NM and W). We have:
  • 300 RPM = 5 rotations per second
  • 8 NM = given value
  • 6.2832 = 2 * pi
  • 0.95 = efficency of the drivetrain
5 * 8 * 6.2832 / 0.95 = approx 265W riders input (this is a high value for an ordinary rider, but the calculation is fine)

The key point is: Why do we have 300 RPM?

It's because we have the rider's power and the motor's power together to achieve 300 RPM. And what will happen if we reduce the power of the motor but the rider still contributes with the same power? We will have e.g. 240 RPM. But as mentioned, the rider makes stil it's 265W. The formula is saying now: 4 * 8 * 6.2832 / 0.95 = approx 265W riders input. So, the formula claims the rides has reduced input -but this is not true, as we said, he maintains his power. We only reduced the contribution of the motor.

The reduction of the RPM can also occur, if the demand rises. E.g. if the wind increases or a climb is there. Then motor and rider contribution still with the same power values, but the speed (=RPM) decreases.

The RPM are not a given value, its variable which can change independently from the other values in the equation.

You see the dilemma I'm in?
 

tbar23

Member
“The RPM are not a given value, its variable which can change independently from the other values in the equation.”
Ah @bluecat but it is a known value.
It is a direct drive motor so RPM is locked to bike speed (unless the rear tire is slipping).
In the case of the ST2S, the 26” x 2.0” tires have a diameter of about 2073 mm = 2.07e-3 km
2.07e-3 km x 300 rpm x 60 min/h = 37.3 kph (23 mph)

So the calculation that @Johnny provided above says that 8 N-m of torque at the axle of a Stromer going 37.3 kph corresponds to ~250 W.
At half that speed, 18.6 kph / 11.5 mph, that same 8 N-m of torque corresponds to half as much power, 125 W. 250 W would require 16 N-m of torque.

The speed sensor provides RPM And the torque sensor provides torque. Everything needed to calculate power.

Let’s take your first example. Rider power plus motor power achieve a bike speed corresponding to 300 rpm motor speed with the rider providing 250 W of power. What we haven’t said is what the motor power is. Motor power will be related to the cross-sectional area of the bike+rider, the wind speed, the road gradient, somewhat related to the tires, whether the bike+rider is accelerating, at constant speed or decelerating.
If the bike+rider is maintaining constant speed, then motor power + rider power must simply balance all losses (aerodynamic drag, drivetrain mechanical loss and rolling friction). At 37.3 kph, and a flat grade, loss is almost certainly dominated by aerodynamics.
Let’s add some details to the example:
assume that the rider is maintaining a constant speed of 37.3 kph (300 rpm) by providing 250 W with the motor also providing 250 W while riding on a 2%, smooth grade and a 12 kph head wind.
If this is all true, then 500 W is required for this bike+rider combo to maintain a steady speed on the example grade with given wind to balance all loss.
If motor power is reduced and nothing else changes, then the bike+rider combo will decelerate to a new speed where 250 W plus new motor power, say 150 W, = 400 W balances all loss. You can find several online calculators to help estimate this scenario (Gribble bike calculator or Bike Calculator)
 
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bluecat

Well-Known Member
The speed sensor provides RPM And the torque sensor provides torque. Everything needed to calculate power.

The original claim was: A Stromer has all aboard to calculate the riders power.

This is - if it comes to e E-Bike with hub drive - wrong. The lack is, that you assume with changing RPM the riders input power also changes. You're assuming this because of the formula. The formula is correct, but not usable in this case. In fact, the riders power can be always the same regardless of the speed. It might be easier to see in a practical example:

Find a e-bike rider who has a real power meter (measuring of force and cadence at the crank). Ask him to ride with 100W own power. Now ask him to change the assist levels, but keeping the 100W own power. As a result, he will be faster or slower - and with the formula above, his power will be increase or decrease. But it might be difficult to find such a rider for a test. So I propose a self-test with your Stromer:
  1. Choose a wide open and save road with constant conditions.
  2. Set level 2 / snow.
  3. Ride at the upper end of your comfort zone, so breath- and heart rate are very stable.
  4. Now switch to level 3 and maintain heart- and breath rate. This shall simulate constant power from your side.
  5. Question: Does the speed (=the RPM) change?
Did my words bring some clarification?
 

tbar23

Member
"Now ask him to change the assist levels, but keeping the 100W own power. As a result, he will be faster or slower" - correct. If assist is increased, rider will be faster. If assist is decreased, rider will be slower - holding all other variables constant.
"and with the formula above, his power will be increase or decrease." - no!
This should be corrected to read: "and with the formula above, total power will increase or decrease.", where total power is the sum of the eBike motor power and the rider power.
To separate the motor power from the human power, you also need to measure motor current and voltage - something Stromer almost certainly does.
The big assumption here is that Stromer has characterized the conversion of motor electric power to mechanical output power, but given the existence of threads like this: eBike motor efficiency I'm sure they are well aware of the performance of their motor at different input powers and different hub speeds.

How accurate would this be? Who knows, but all of the information is available and/or can be estimated with reasonable accuracy (e.g. motor efficiency).
 

bluecat

Well-Known Member
to separate the motor power from the human power, you also need to measure motor current and voltage

Great, you did the turn to the right direction. As earlier mentioned, you need also to know the contribution of the motor:

  • You have the speed, which is the result of the riders and motors energy
  • You need to have the motors energy consumption
  • You need to have the motors efficiency at the given speed
  • You have the riders force, translated by the drivetrain
  • You need to have the drivetrain's efficiency at the given gear
I would say, with a bucket full of assumptions, Stromer could bring a "Riders Power" value to the display. For entertainment, with a two decimal precision.

Unfortunately, I've omitted something. With all these informations, you'll get the riders power only under standard conditions. I have to come back to this:
The key point is: Why do we have 300 RPM?

On a flat road with no wind, you might get a value for the riders power which is more or less realistic. Luckily, you provided this link, which allows us a simulation:
You can find several online calculators to help estimate this scenario (Gribble bike calculator
  • Use the default values and set 100W rider power
  • Look what speed is calculated (25.4 km/h)
  • Now add 200W power (to be the motors contribution), in total, you have now 300W
  • Look what speed is calculated (39 km/h)
  • Now set a climb of 5%
  • Look what speed is calculated (20.7 km/h)
With other words: From speed ( =RPM), you cannot calculate the riders power. You cannot do this even if you have the exact contribution of the motor.
 

tbar23

Member
@bluecat your final analysis is not correct.
If the speed drops from 39 km/h to 20.7 km/h and the total power (300 W = 200 W motor plus 100 W rider), then the torque must increase. So the RPM will have dropped by 20.7/39 = 53% of its previous value and the torque must increase by that same amount. The TMM sensor will measure this and know that the total power is still 300 W independent of the grade or the wind or whatever.
If TMM measures a torque lower than this at the lower RPM, then the motor + rider are no longer producing 300 W.