Good news for me: Briggs and Stratton will provide torque curves and related data if you ask them.
So for the 40H777-0241-E1 engine used on the GT2254 riding mower from 2005, I received the following table:
|Oil Capacity||64 oz (1.9 L)|
|Flywheel Torque||130 lb/ft (176.0 Nm)|
|Governor Arm Torque||70 lb/in (8.0 Nm)|
|Carb Mount||6.5 lb/in (7.5 Nm)|
|Plug Gap||0.03 in (0.75 mm)|
|Intake Valve||0.004- 0.006 in (.10 -.15 mm)|
|Exhaust Valve||.004 - .006 in (.10 - .15 mm)|
|Idle Speed||1750 RPM|
|Air Gap||.008 - .012 in (.20 to .30 mm)|
|Flywheel Puller Part||19203|
|Flywheel Holder Part||19433|
|Connecting Rod Torque||100 lb/in (11.5 Nm)|
|Sump Torque||200 lb/in (22.5 Nm)|
|Head Torque||220 lb/in (25.0 Nm)|
Husqvarna also responded for a request regarding the load ratings of the lawn mower in question, stating that it was rated for 846 lbs of No Load Weight. This seems illogical to me, as I highly doubt the mower weighs this much, even with the deck installed, and somebody much heavier than myself riding on it. Full Load data was not provided, and when asked for clarification, I was told that the 846 lbs load was specifically No-Load.
HP of a motor is measured at full RPM, but Torque ratings are taken at 3060 RPM according to a gross power chart that came with the table above. And do not include losses from installing the air cleaner or exhaust systems. This makes things slightly trickier, but I'll worry about that loss later.
Horsepower = Torque * RPM / 5252
(Source page 1, column 2)
This source also gives Mechanical HP = 746 W = Electric HP.
Torque is defined by science as kg * m^2 / s^2 in units. Or mass * distance^2 / time^2.
The ratio between the drive shaft torque and the flywheel torque is simple to calculate if I treat the flywheel and shaft as one assembly. This causes the mass parts of the Torque definition to cancel itself out, and since they're both acting within the same timespan, the time variables cancel out. Leaving a ratio of two circumferences. The diameter of the shaft is listed at 1.125 inches and measures within tolerance of that value. The diameter of the flywheel, I am taking from measurements as well, and comes out to 9.5 inches (approx.). Thus:
T_flywheel / T_shaft = ( pi() * 9.5 )^2 / ( pi() * 1.125 )^2 = C_flywheel^2 / C_shaft^2
~Remember, diameter is 2x radius, and circumference (C) is 2 * pi * r .~
or T_flywheel / T_shaft = 890.732 / 49.965 (within about a significant figure of tolerance)
Thus: T_flywheel / T_shaft = 176.0 Nm / T_shaft = 890.732 / 49.965
T_flywheel * C_shaft^2 = T_shaft * C_flywheel^2
T_flywheel * C_shaft^2 / C_flywheel^2 = T_shaft
176.0 * 49.965 / 890.732 = T_shaft
8793.8 / 890.7 = 9.9 Nm within one sig fig of tolerance. (slightly less if ignoring significant figure truncations)
This is what I have for today's update. I need to do some research into the motors and such before I continue further.