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#1
October 27th 06, 07:31 PM
posted to rec.aviation.ifr
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12 horsepower per hour per gallon, equates to percent power
Rule of thumb. So if you are burning 10 gallons an hour in a 180 hp
engine, you are developing (10x12) = 120 hp. 120/180 = .67, so that is 67% power. It's not exact but gets you close. 
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#2
October 27th 06, 11:23 PM
posted to rec.aviation.ifr
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12 horsepower per hour per gallon, equates to percent power
Doug wrote:
Rule of thumb. So if you are burning 10 gallons an hour in a 180 hp engine, you are developing (10x12) = 120 hp. 120/180 = .67, so that is 67% power. It's not exact but gets you close. How does altitude get involved? According to the chart in the Lycoming Operator's Manual, full throttle HP at Altitude there's roughly a 3% loss of HP for every 1000 ft. above sea level.
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#3
October 28th 06, 12:15 AM
posted to rec.aviation.ifr
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12 horsepower per hour per gallon, equates to percent power
"Blanche" wrote in message ...
Doug wrote: Rule of thumb. So if you are burning 10 gallons an hour in a 180 hp engine, you are developing (10x12) = 120 hp. 120/180 = .67, so that is 67% power. It's not exact but gets you close. How does altitude get involved? According to the chart in the Lycoming Operator's Manual, full throttle HP at Altitude there's roughly a 3% loss of HP for every 1000 ft. above sea level. Assuming proper leaning and unchanging RPM, falling manifold pressure decreases fuel flow. Then, Doug's relationship remains independent of altitude. It's derived from a simple statement of avgas' energy content. It just assumes the engine converts chemical energy into go-power. As Doug mentioned it's approximate, but useful.
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#4
October 28th 06, 06:11 PM
posted to rec.aviation.ifr
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12 horsepower per hour per gallon, equates to percent power
Doug wrote:
Rule of thumb. So if you are burning 10 gallons an hour in a 180 hp engine, you are developing (10x12) = 120 hp. 120/180 = .67, so that is 67% power. It's not exact but gets you close. Sure, but unless you have a fuel flow gauge, you don't know how much fuel you are burning, and you would have to resort to rpm and density altitude to figure out the % power.
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