The student returns with another question![]()
Can someone please check my math?
We did calculations for Flywheel Moment of Inertia.
I = (k)mr^2
Sub-divided each part, used the mass for that part, and the k constant that made sense for that part; k=1 for the rim, k=0.5 for the hub, etc.
Using pounds for mass and feet for radius, this is how we got to 266 ld ft^2 for the flywheel by itself.
Using the same units for the drop test, we ended up with 280 ld ft^2.
Okay, now I have a value for Inertia in the torque equation T=Ia
All bright eyed and bushy tailed, I calculated the angular acceleration of the flywheel for a test pull that was just conducted with a 212cc 4 stroke engine that should make 8.1ft ld ft of torque peak, and about 6.5 hp peak. (Honda Generator Engine Clone)
I dont have the recorded pull data to see the actual curve shape, but I know the start and end engine RPM's of the pull, I know the time of the pull in seconds, and I know the gear ratio between engine and flywheel shaft.
RPM range = 1000 Engine RPM to 5000 Engine RPM
Pull Time = 15 seconds
Gear Ratio = 6:1 (10t on engine crankshaft, 60t on flywheel, no transmission, no clutch)
Thus, averaged across the pull, the engine accelerated at a rate of 266.66 RPM's per second.
Using the 6:1 gear reduction, this means the flywheel accelerated at an average rate of 44.5 RPM's per second.
The software has a sampling frequency of 5 samples/second.
So, I decided to look at 0.2 second "steps" of time for the rate of angular acceleration (a). Using the average acceleration rate of the Flywheel I whitled down to:
44.5 RPM's per second = 0.74 RPS's per second = 0.148 RPS's per data point (per 0.2 seconds) = 0.93 angular velocity (w) per data point
So, if this is averaged across the whole run, each data point will have the same (w), and thus the same delta(w).
Same thing for time, every data point is 0.2 seconds "wide", so delta(t) will always be 0.2.
Finaly;
a = delta(w) / delta(t)
a = 0.93 / 0.2
a = 4.65
Now, armed with all of the mathematical powers needed to divine the measured engine power.... everything falls appart.
T=Ia
T=280 * 4.65
T=1302
1302 ft lds of Torque measured at the flywheel shaft?!?!?!
-okay, thats a little big, but thats flywheel shaft torque, so lets divide it by the 6:1 gear ratio to find out what the engine did....
1302 ft lds / 6 = 217 ft lds of "average" Engine Torque across the 15 second pull.
Thats a fever dream, not a measurement....
After considering my life decisions for a few hours, I think I stumbled into my problem:
I = (k)mr^2 only works with SI units.
The output of that formula when we use pounds and feet is absolute bunk for executing the T=Ia formula.
So, I redid the Moment of Inertia Calcuation using SI units.
I end up with 11.8 kg m^2 after converting the weight of the flywheel in lds to kgs (110ld flywheel), and converting the radius to meters (2ft radius flywheel).
-lets try this one more time:
T=Ia
T=11.8 * 4.65
T=54.87
54.87 Nm of Torque measured at the Flywheel shaft, that sounds more like reality.
54.87 Nm lds / 6 = 9.15 Nm of "average" Engine Torque across the 15 second pull.
9.15 Nm converts to 6.75 ft lds.
6.75 "average" ft lds across a pull from 1000 - 5000, using a 4 stroke engine that makes ~8.1 ft lds of peak torque at 2,500 RPM?
-considering this 6.75 ft lds is "uncorrected power", and the ~8.1 ft lds number is the manufacturer advertised, SAE corrected output.
That sounds like a real measurement.
Now, lets check HP using flywheel RPM like Neels just advised.
To do this, I took the 54.87 Nm of flywheel shaft torque and converted it to ft lds;
54.87Nm converts to 40.47 ft lds
Using 40.47 for torque and flywheel RPM for RPM, I calculated HP at 1000 Engine RPM (166 flywheel rpm) and 5000 Engine RPM (833 flywheel rpm).
HP = T * (RPM / 5252)
If the Torque curve is averaged to the "flat line" of 6.75 ft lds of torque at every RPM step like I have done;
HP at 1000 RPM = 1.3 HP
HP at 5000 RPM = 6.4 HP
That is "in the ballpark" as they say.
Final sanity check;
The pull did not go to 5252 RPM, but if I keep extending the averaged 6.75 ft lds of torque out to 5252;
HP at 5252 RPM = 6.75
Torque at 5252 RPM = 6.75
HP and Torque "cross over" at 5252 RPM's!
Woohoo!
Can someone please check my math and give me a positive confirmation that I = (k)mr^2 only works with SI units for the purpose of using in the formula T=Ia?
Bookmarks