longwinglover
Well-Known Member
- Joined
- Dec 28, 2004
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Got a question for all you guys out there that are smarter than me (that may be ALL of you ;D ).
Here are some what I will call "givens"
Horsepower = (Torque x RPM) / 5252. I'm not sure that THIS formula is correctly applied here.
"Standard conditions" i.e. Sea level and 59*F for on the ground and +50ft for in the air.
Continental O-200 engine - 100hp at 2750 RPM.
With a properly pitched prop the engine will give a static RPM of 2750 (100hp) at full throttle
(Wide Open Throttle - WOT). I assume this is also 100% torque???
The question:
If you take off and fly along the beach at 50ft and stabilized airspeed (if it matters let's say 100mph) for 2750 RPM. How much HP is the engine making? Is there a simple formula for this? Is it incalculable with the information given? Why do I care?
I know that the throttle is no longer WOT because I've had to pull it back (close the butterfly) to keep the RPM at 2750. I assume the Torque has also reduced due to a) reducing the throttle and b) the airflow through the prop due to the movement through the air (100 mph airspeed).
Can the formula be worked "backwards" from 2750RPM at 100mph at WOT to figure HP produced at "X" static RPM WOT on the ground?
John Scott
Here are some what I will call "givens"
Horsepower = (Torque x RPM) / 5252. I'm not sure that THIS formula is correctly applied here.
"Standard conditions" i.e. Sea level and 59*F for on the ground and +50ft for in the air.
Continental O-200 engine - 100hp at 2750 RPM.
With a properly pitched prop the engine will give a static RPM of 2750 (100hp) at full throttle
(Wide Open Throttle - WOT). I assume this is also 100% torque???
The question:
If you take off and fly along the beach at 50ft and stabilized airspeed (if it matters let's say 100mph) for 2750 RPM. How much HP is the engine making? Is there a simple formula for this? Is it incalculable with the information given? Why do I care?
I know that the throttle is no longer WOT because I've had to pull it back (close the butterfly) to keep the RPM at 2750. I assume the Torque has also reduced due to a) reducing the throttle and b) the airflow through the prop due to the movement through the air (100 mph airspeed).
Can the formula be worked "backwards" from 2750RPM at 100mph at WOT to figure HP produced at "X" static RPM WOT on the ground?
John Scott
