12 horsepower per hour per gallon, equates to percent power

[Doug[_1_]]

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.

[Blanche]

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.

[John R. Copeland]

"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.

[Andrew Sarangan[_1_]]

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.