If gears multiply torque then why does a Dyno read less?

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Discussion Overview

The discussion revolves around the relationship between gear ratios, torque multiplication, and the readings obtained from a dynamometer (dyno). Participants explore why a dyno may report lower torque values than expected based on calculated torque from gear ratios and engine specifications.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant questions why a dyno reads close to the engine's original torque value despite calculations suggesting a higher torque due to gear ratios.
  • Another participant notes that dynos typically determine power first, referencing the physics involved in dyno measurements.
  • Some participants suggest that the dyno calculates torque based on traction force and speed, with the possibility of back-calculating engine torque if gear ratios and tire diameter are known.
  • There is a discussion about the first law of thermodynamics being applicable to relate power at the tire patch to engine power, with considerations for mechanical losses in the system.
  • One participant mentions that without knowing gear ratios, the dyno can still provide torque estimates using power balance equations, factoring in mechanical efficiency.
  • Concerns are raised about how to determine torque for vehicles with unknown gear ratios, with suggestions for measuring engine RPM and calculating drive train ratios from speed and wheel radius.

Areas of Agreement / Disagreement

Participants express varying viewpoints on the mechanics of dyno readings and the implications of gear ratios on torque calculations. There is no consensus on the exact reasons for the observed dyno readings, indicating ongoing debate and exploration of the topic.

Contextual Notes

Participants highlight limitations in the accuracy of torque calculations due to potential errors in measuring RPM and speed, as well as the assumptions made regarding mechanical efficiency.

knight92
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If the selected gear ratio and final drive ratio multiply the torque and reduce the speed then why does the dyno read less torque?

e.g a car with 200 Nm (@5000 RPM) of torque with a 2nd gear selected (Ratio = 2) and final drive ratio 3.5

200*2*3.5 = 1400 Nm of torque
5000/2/3.5 = 714.3 RPM

Why would the dyno read close to 200 Nm then? is it the corrected torque?
 
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knight92, you didn't mention it, but it sounds like you are measuring your output on a chassis dynamometer driven by your tires. In that case, the dyno only "knows" two things about your car: the traction force exerted on the dyno rolls (in Newtons, say), and the speed at the tire patch (in meters/sec, say). But if you type your gear ratios into the controller, and the tire diameter, the software in the controller can back-calculate the torque at the engine (assuming some level of friction in the driveline). That may be what it's doing.
 
Randy Beikmann said:
knight92, you didn't mention it, but it sounds like you are measuring your output on a chassis dynamometer driven by your tires. In that case, the dyno only "knows" two things about your car: the traction force exerted on the dyno rolls (in Newtons, say), and the speed at the tire patch (in meters/sec, say). But if you type your gear ratios into the controller, and the tire diameter, the software in the controller can back-calculate the torque at the engine (assuming some level of friction in the driveline). That may be what it's doing.
Thats what I was wondering whether the torque shown is back calculated but then not every car's gear ratios are known so how can someone put a random car on the dyno and know what torque the engine is delivering.
 
knight92, if you don't know the gear ratios, you can use the first law of thermodynamics (or at least the time-derivative of it) by saying "power at the tire patch equals power from the engine." (Of course the engine power must be more, because of the mechanical losses in-between.) So you can write force X velocity = engine torque X engine angular velocity, that is, Ptire = Pengine. The dyno should either read out force (surface force at the tire patch) and velocity, or power at the tire patch (Ptire). You should also be recording engine speed, so at that point it's just engine torque = Ptire/omega, where omega is the engine's angular velocity.
To be more proper, you would multiply the engine power by eta, the combined mechanical efficiency of the driveline and tires, which is somewhere around 0.85. Using that, the engine torque would be Ptire/(omega*0.85), increasing the back-calculated engine torque by about 18%.
Note that without accounting for the mechanical efficiency eta, the engine torque predicted from the power at the tire patch would be less than the engine's actual torque, just as you originally said!
And you're right, the more general way to do this is from a power-balance standpoint. Then you don't have to worry about gear ratios, or even having a torque converter in the system.
 
knight92 said:
Thats what I was wondering whether the torque shown is back calculated but then not every car's gear ratios are known so how can someone put a random car on the dyno and know what torque the engine is delivering.
In the unlikely event you can't find ratio data online, Measure the engine rpm (at the dizzy or via ecu) : engine rpm over dyno rpm gives drive train (gear box and diff) ratio.
You can also calc drive train ratio on the open road from the rpm, speed and wheel radius. Error in your tacho and speedo mean it won't be very accurate through, I'd guess ~5% for an average car.
 
billy_joule said:
In the unlikely event you can't find ratio data online, measure the engine rpm (at the dizzy or via ecu).
or simpler still, simply attach a sensing wire / clamp to one of the spark plug wires (this is done at some dyno shops).
 
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