Quote:
Originally Posted by Captain Snooze
@ stugray
This question isn't aimed at this thread in particular; it is a physics question. I can't remember too much from my study days 
If one is using an inertia dyno is one actually measuring power or is power calculated?
That is, a known mass is accelerated in a known time; is this force,work or power?
Ta. |
In this case (simple inertia dyno) the only things that are measured are Drum RPM, and time.
Using those two numbers (drum angular velocity) and elapsed time (
∆t), you can derive the power.
Power is the rate of doing work.
The work done is the change in rotational energy of the drum.
If you start from a stop, w = 0 so rotational kinetic energy is zero
The power is the final kinetic energy of the drum (in joules) divided by the time it took to spin the drum to the final speed.
So rotational kinetic energy is E = 1/2*I*w^2 ( where E is in joules, I is drum rotational Inertia in kg-m^2, and w is angular velocity in radians/second).
So total energy is: (Efinal - Estart)/time-to-reach-Efinal (in seconds).
This will give an average "Joules per second" which is the definition of Watts.
Note: this will only give you average power over the whole "pull" and will not provide the nice dyno charts we typically see which represent instantaneous power across the RPM range.
This precisely how I calclulated an average hp draw of spinning just the driveshaft.
for the driveshaft example:
Rotational inertia of the 30 lb driveshaft (@3 inch diameter) is : 0.01974 kg.m^2
wstart = 0 RPM, wfinal = 5600 RPM
5600 RPM = 35185.8 radians per minute = 586 radians per second
The energy stored in the driveshaft = E = 1/2 * .01974*586^2 = 3389 joules
If it took 16.2 seconds to spin the driveshaft from zero to 5600 RPM, then the 3389 joules were spread out over 16.2 seconds so the power (energy per unit time) is: 3389 joules/16.2 seconds = 209.2 watts.
209 watts = .28 horsepower.
Side note: So since the
headlights consume more than 200 watts, I would argue that driving with your headlights off vs On, would make more of a difference than removing the entire 30 pounds from the driveshaft (NOT accounting for the reduction
in overall vehicle weight). SO I am sure to start whole new flurry of naysayers with that one.....
So to be clear which of the following experiments would make the bigger difference:
Replace the stock driveshaft with a 15 pound lighter one (shed 15 pounds from vehicle weight AND 15 lbs from rotational mass at the driveshaft) and do a 1/4 mile run with the headlights ON
AND
Remove 15 pounds from the spare tire (shed 15 pounds of vehicle weight) and do a 1/4 mile run with the headlights OFF.
Which run is faster?
The one where you
removed the weight from the spare tire, and drove with headlights off.