Exactly....but you cant make a drum that large and fill it with concrete or whatever is in it to extremely close tolerance so that one calibration fits every drum. I am suspecting they have something back at the factory that applies a known Tq to the drum to calibrate each drum. My point is you cant DIRECTLY measure HP like you can TQ. Even the dynojets HP measurement is a calculation of HP based on time and acceleration which in turn was calibrated back at the facory with DIRECT MEASUREMENT of tq applied to that specific drum.

 

I am refering to an inertial dyno (dynojet) not a load dyno (mustang dyno).

You can't just measure the torque at the wheels and use the engine RPM for the calculation. The numbers would change dramatically with gear ratio differences, and they don't. What is really measured is horsepower at the rear wheels and that then is translated into engine torque via the normal calculation based on engine rpm.

Here is a great explanation that I found (I would claim it as my own, but everyone here knows I ain't no damn mathmagician!):

"A combination of two laws of physics, force equals mass times acceleration and work equals force times distance, gives us this equation: W=m X a X d. "W" is the work, in pounds-feet, the rear wheels are doing, "m" is mass equivalent (the drums), "a" is acceleration (increasing drive wheel speed) and "d" is distance (drum circumference). Once we have the work, we can find horsepower. One horsepower is 550 pounds-feet of work done in one second so, we divide the work number by the length of time measured, then divide the number we get from that by 550. To simplify: we get horsepower by multiplying the mass, acceleration and the distance, then dividing that product by time multiplied by 550. This can be expressed by: hp = (m X a X d) ÷ (t X 550).

Torque can be figured by multiplying the horsepower by a constant, 5252, then dividing that product by the speed at which the thrust force was measured. Generally, with rear wheel numbers, axle ratio is not considered in the torque computation. For comparison purposes, this makes more sense. The computer factors out the axle ratio by using engine speed data in the torque derivation."

I know that is sorta hard to follow but if you really think about what is saying you can see that rear wheel horsepower is measured before engine horepower OR torque is calculated...thus eliminating the gear ratio factor.

 

The key point im trying to make is DIRECTLY MEASURE. there is no machine anywhere that I know of that can directly measure HP. Everything used to determine HP will be a calculation of atleast two parameters being directly measured. Wether those two parameters are time and mass or TQ and rpm. There will always be two. Tq on the other hand...is just measured straight up right from the source...DIRECTLY.

 

That pretty much sums it up Dynojet won the advertising and sales pitch wars big time. Dont get me wrong though...I like the dynojet.

 

the dynojet measures HP indirectly as its primary goal.

it determines HP based on the change in angular velocity of a known mass of a steel drum. in theory the dynoject measures nothing. it calculates time based on a processor operation. then it calculates the amount of time between speed sensor puleses off the axle. it then knows (by programmed value) what sensor pulse rate equals what angular velocity. using a known mass and moment of inertia of the steel drum assembly the computer is able apply theses values to a formula. if the cpu can determine how much the angular velocity of a known mass has changed over a given period of time it can then determine how much work is being done to the mass and thus how much power is being delivered.

so...

there is nothing in a dynojet that MEASUERES tq. the tq is calculated by taking the calculated HP and dividing it by an RPM value that the sensor box acquires off the vehicle.

if you make a pull on the dynojet and dont hook up any sensors to the vehicle you will get a graph of HP vs. wheel speed! so no TQ is measured.

whoever stated that all dynos measure TQ is wrong ...and i know the british website that you got that quote off of...and he is wrogn in that statement.

the Dynojet does not measure TQ or HP. it measures time..and then calculates HP.

 

I'm pretty comfortable with this. It makes perfect sense to measure the changes in velocity over time of a known amount of mass to calculate HP. And its probably much easier than applying a constant resistive force, such as a brake or an electric motor functioning as a brake (generator mode).

I was under the impression that the mass type dyno's mask rapid fluctuations in the HP/TQ curves. I can't remember the exact name, but in the threads where a guy has the 'ultimate garage', he had a dyno system that attached directly to the hubs.

 

my question is, if f = m x a (which obviously it does), what force f is being measured? torque? torque divided by the wheel's radius, as mentioned above? and is that an oversimplification? obviously it's negating outside factors like wind resistance, etc. but as far as the actual mechanics involved, what is the force which, when divided by the mass of the vehicle (or would it be the mass of the engine?), gives you your acceleration?

and did any of that even make sense?

 

you are correct.

the driving "force" accelerating the vehicle (which i shall call thrust) is the force the road exerts on the tires (which is equal in magnitude and opposite in direction to the force the tires exert on the road--newton's third law thing).

basically as the wheel turns (via torque from the axle), it exerts a force on the road tangental to the surface of the tire (that is, perpendicular to the RADIUS of the wheel). the torque the axle delivers is determined by the torque the engine produces, multiplied by the mechanical advantage of the gearing, and subtracting any frictional losses through the drivetrain.

the car is accelerated by the force the road exerts on the tire, and this is equivalent to the torque the tire exerts on the road (again, by newton's 3rd law), divided by the radius of the tire (which is equivalent to the force vector portion of the torque).