Dynamometer and torque questions

[eagle]

We were discussing dynos the other day and it brought up some interesting questions. Hopefully there are some math guys here to help answer this.

 

A bike engine puts out say 100HP and 50lbs of torque at the crank. Running on a chassis dyno, the final gear ratio would be 4:1 ( 4 crank revs to 1 tire rev). So we would multiply the torque by 4 at the rear wheel so 200ft/lbs at the rear tire.

 

Here is one thing I do not understand, how much torque does the dyno roller "see" if the roller is 16in diameter and the tire is 20in? Did I miss a step? Do I just need to compare the tire diam to the roller diam?

 

 

Also, when I see a dyno chart, it shows say 100HP and 50lbs of torque at the back tire, what does that mean and how is it calculated? I mean we are seeing far more than 50 lbs at eh back tire. Is this just somehow re rated just like the tire above?

 

:shrug:

[BSHEE400]

A bike engine puts out say 100HP and 50lbs of torque at the crank. Running on a chassis dyno, the final gear ratio would be 4:1 ( 4 crank revs to 1 tire rev). So we would multiply the torque by 4 at the rear wheel so 200ft/lbs at the rear tire.

 

 

i'm no expert on dynos, not by a long shot.

but that sounds way off to me, i'm under the impression that crank dynos will always show higher numbers than a chassis dyno because you lose power through the drivetrain and the weight of wheels and tyres etc.

also on the final gear ratio, i think it refers to the ratio of SPEED not power but i could be wrong.

 

someone with more detailed and accurate info will chime in soon

[desertdeweller]

yes you will see less at the tires however I do not know how the ratios work out either. I think you typically have like 15 to 25% loss at the whells compared to the crank.

[moneybags]

A bike engine puts out say 100HP and 50lbs of torque at the crank. Running on a chassis dyno, the final gear ratio would be 4:1 ( 4 crank revs to 1 tire rev). So we would multiply the torque by 4 at the rear wheel so 200ft/lbs at the rear tire.

 

Also, when I see a dyno chart, it shows say 100HP and 50lbs of torque at the back tire, what does that mean and how is it calculated? I mean we are seeing far more than 50 lbs at eh back tire. Is this just somehow re rated just like the tire above?

:shrug:

I can tell you this much. Your bike does not have 200 pounds of torque.

[eagle]

I can tell you this much. Your bike does not have 200 pounds of torque.

 

 

 

The numbers represented above are just test figures but yes, simple physics applied will show indeed that 50lbs of torque going through a 4:1 ratio gear reduction, will produce 200lbs of torque but at a relative rpm. This does NOT however, indicate what will be represented on a dyno graph because those numbers are calculated. I would like to know the math behind this and if the final gear ratio is figured back in to determine a calculated torque value.

 

 

Tell me how a 50hp tractor is able to pull a house down? Gear reduction as a torque multiplier.

Edited February 24, 2009 by eagle

[LS1Inferno]

this would be a royal PITA to calculate BUT when i get home and have them time to do it i will if you have 50Lb ft of tq i could tell you how much you are making at the output shaft. the has to do with the ration but not at 4:1 thats where rpm and HP comes from. you have 50 lbft at a wheel with a 10"radius which and tq = F*Distance do 50/(10/12)= 60lbs of force on the axle and then you need the radius of both sprockets. you could calculate it all the way back to the crank if you know all the gear sizes its a long process. but usually about 5-10% loss is my guess we use 20% drivetrain loss in cars

[eagle]

this would be a royal PITA to calculate BUT when i get home and have them time to do it i will if you have 50Lb ft of tq i could tell you how much you are making at the output shaft. the has to do with the ration but not at 4:1 thats where rpm and HP comes from. you have 50 lbft at a wheel with a 10"radius which and tq = F*Distance do 50/(10/12)= 60lbs of force on the axle and then you need the radius of both sprockets. you could calculate it all the way back to the crank if you know all the gear sizes its a long process. but usually about 5-10% loss is my guess we use 20% drivetrain loss in cars

 

 

The method I used to derive the rough 4:1 ratio (which is not precise) is by comparing a top speed to rpm, and the tire size. If you know a stock on stock banshee does 78mph at 10K with 20 tires, you can determine the final axle ratio including all gear reductions at the crank, trans, and chain sprockets. Obviously I have a more accurate way of doing this but this is just to better understand the math.

 

What I see is say 200ft/lb of force at the back axle/tire size in ft to determine moment force on roller. This works out to 200/(20/12) or 119.8lbs of force.

 

What I am getting at is all these damn inertial dynos floating around testing power but without proper ratios in place, it is a GUESS! I saw this on another dyno console that said "standard calculation", asked the operator what it was, he said, "dunno but it works". Bottom like is how the hell can they be anywhere close to accurate?