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topless -
your calculations aren't right. you're just assuming a 3% loss for each 1000'. the pressure drop is NOT linear.
the SAE correction equation is:
cf = 1.176 * ((990/Pd)*(Tk/298)^.5)-.176
one over this cf yields the percent power available at that pressure. Pd is the pressure of dry air at altitude, and Tk is the temperature in degrees kelvin.
i'll use 15 degrees C since that's what my altitude pressure tables are calculated for. at 5,500', the pressure of dry air is about 820mb.
if i plug all this into the equation, i get a cf of 1.22 or an HP yield of 82%.
that means that if your assumption of 260RWHP at sea level were used, the HP available at 5,500' asl is 213.
FYI, Raby says he usually sees about 240RWHP from the 3.4L. this would yield 196RWHP at that altitude.
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