</font><font color="blue" class="small">( </font><font color="blueclass=small">( The only thing that matters is what rpm you're measuring the torque at. If they both make
10HP@10RPM, they both have 5,250 ft/lbs of torque.)</font>
This is on the right track, but presented a little backwards, since the inital equality was HP, so what matters is at what RPM the HP was measured (semantics).
Torque is a real measured performance parameter, HP is simply what falls out of an equation, as HP = torque*RPM. Torque can be viewed as proportional to how much oomph a single firing of a single piston makes. Ideally, torque should remain more-or-less constant over a usable RPM range (pretty much true for diesels at least) so HP should ramp proportionally with RPM.
So, not enough information on the initial question. What are the displacements? More than likely, with a 3 cylinder engine vs. a 2-cylinder of the same HP rating, the 3-cylinder will probably be able to spin faster (since the pistons are likely smaller) and therefore make the same HP with less torque, but at a higher RPM. OTOH, the 3-cylinder could be a slow-revving torque monster like a diesel tractor motor, while the 2-cylinder could be a fast-revving teeny-torque motor with all of the HP made with RPMs, like a crotch-rocket motorcycle engine.
So, it could go either way - for example, my 4-cylinder Kubota
L4200 makes 42HP (depending on who is doing the quoting) at ~2000RPM or so. A single-cylinder Honda CR250R two-stroke dirt bike motor also makes 42HP, but has to scream to do so.
Jay )</font>
Torque is a force, measured as ft/lbs in this case. Horsepower is a measure of how much work that force can do over a period of time. The actual formula is hp=tq*rpm/5252 . The 5252 is an engineering constant.