Dave,
That is a rather open ended question. There are a lot of engine designs out there.
Here are two of them, with some rough calculations of their torque output at the max pressure with a reasonably long cutoff.
[
members.pioneer.net]
Their Double simple engine, 2 1/2" bore by 2" stroke, max pressure 150, max rpm 750.
With a mep of 130(150 psi inlet and .86 expansion factor) and a card factor of .75 it would be 25 ft lbs of torque per cylinder or 50 for both cylinders. That is the average torque per revolution, the actual stall torque will depend on what crank angle there is and if only one or two of the cylinders are getting steam.
If you could supply that much steam at 750 rpm that would be around 7 hp, a bit more.
[
hasbrouck.8m.com]
This one is a V twin, 1.875" bore by 2" stroke. 400 rpm max, 120 psi.
This with more or less the same parameters as above would make around 11 ft lb of torque per cylinder or 22 from both.
Now the Reliable engine will weight 87 lbs and the Hasbrouck will weigh 48 lbs. Neither of these are light.
A conservative figure for rolling resistance is 35 lbs per 1,000 of vehicle weight on grass or dirt, depending on tire size and width etc.
For a grade it equals, (n*W)/100 = resistance in lbs.
n: grade percent
W: vehicle weight
So for a 300 lb bike and rider weight on the flat that would be 10.5 lbs.
For a grade of 10% 30 lbs added, so 40.5 lbs total.
Figure the total gearing from the engine to the wheel, multiply the engine torque by that, then divide that by the tire radii in feet, that will give you the tractive effort. One should actually multiply the tractive effort by the percentage of friction losses involved between the engine and wheel also to get a more acurate figure.
Lets say the bike has 26" tires, that would be a 1.0833 foot tire radii.
Lets also say that the smallest front cog has 20 teeth and the largest rear cog has 30 teeth. Then the front would spin 1.5 times for each rev of the rear tire. Then also have a 2 to 1 ratio between the engine and the front cog set.
Multiply them together, that is three revs of the engine per rev of rear tire. Then you would need to divide this by 1.0833 to get the tractive effort. Roughly a ft lb of engine torque equals 2.77 lbs of tractive effort.
So on a 10% grade in solid dirt or grass, somewhere around 40.5 lbs of tractive effort are required, that would require 14.6 or about 15 ft lbs of torque from the engine.
On a flat paved road you would only required about 10 lbs per 1,000 vehicle lbs or 3 lbs of tractive effort to get going. That would be just over 1 ft lb of torque.
Note that these forces would just overcome the resistance and not provide much acceleration.
If I were you I would use the foot pedels to get going and then use the steam engine to keep going and accelerate further. That is what mopeds do.
Good Luck,
Caleb Ramsby