If <p>hp = Torque*(RPM/5252)<p>then hp must equal torque at 5252 RPM.<p>This is close on Wett's 2.8 30V graph (assume measurement error).<p>It's way off on their 1.8T graph...they're equal around 4200 RPM.<p>Is there an explanation?<p>steve

just read how a dyno works! measures torque, then the computer attached to the dyno follows the formula to report horsepower.<p>Man...makes the graphs even more inexplicable.<p>steve

Steve,<p>The Wett dyno graphs I have seen have two Y-axes: the left axis for HP, the right axis for torque. The curves are drawn on two different scales, so the torque and HP will not intersect at 5252 rpm. Be sure you are measuring the curve on the correct axis :-)<p>Tom P.<br>

feel like Rosannarosannadanna, "Neee-ver mind"<p>steve

Its easy. 1 Hp is 550 ft*lbf/second. Converting the engine speed to ft/sec uses that ratio known as Pi, 3.141592.... close enough.<br>The engine speed is in revolutions per minute. Divide this by 60 and you get revolutions per second. A 1 foot diameter flywheel would take in Pi feet per turn, jfwiw. (flywheel could be any size) So at 3600 RPM, it would take in 60*Pi feet of rope if it were a sheave (pulley that is a drum) every second. The torque would be twice the pull (force) exerted on the rope as the moment arm is 1/2 foot.<p>If the pull were say 10 lbs, the Hp would be 20 [lbf]*60*Pi[ft/s]/550[Hp/(lbf*ft/s)] or 6.85 Hp.<p>Note the radius or diameter of the sheave doesn't matter as its circumfrence increases the speed of the rope but decreases the force. <p>I use pound as a force unit, hence the lbf. Use dimensional analysis. It got me through my BSME work in thermo, physics and chem.<p>