I've been trying to work out how much extra performance an aircraft will gain over 1st Segment Performance due ground effect prior to climbing to 35'. I've had to make numerous assumptions to guess a performance gain - can anyone look at the following calculations and confirm whether I'm on the right track?
For the purpose of the exercise, I'm using an aircraft with a wingspan of 66' 8", and a wingtip clearance to the ground with no weight on the undercarriage of approximately 12'. The engines are flat rated to 680SHP each, driving 94" four bladed propellers.
I understand the reduction in induced drag due ground effect is 1.4% at 1 span, 23.5% at 0.25 span, and 47.6% at 0.1 span. The undercarriage does not droop at all during the retraction process, and as the gear doors only have to close behind the gear I'd be surprised if there was any significant increase in drag during the retraction process. Using the quarter span wingtip height of 16'8" results in a wheel clearance of just under five feet, which I believe is good enough for the exercise. Accordingly, we can use an assumed induced drag reduction of 23.5%.
To calculate the amount of thrust available, I have used a speed of 89 KTAS, and assumed a propeller efficiency of 0.8. Using the formula thrust in pounds equals HP x prop efficiency x 326 / KTAS gives 680 * 0.8 * 326 / 89 or 1992 lbs for the single engine case. The aircraft has auto feather, so we can discount the drag effect of a windmilling propeller.
I have found several references that show climb gradient equals (thrust - drag) / weight assuming close to horizontal flight. We're going to assume we're operating at weight of 14,000 lbs in conditions that result in a published first segment climb gradient of 0%. As thrust equals drag in this scenario, total drag out of ground effect equals 1992 lbs.
I am going to have to take a wild guess that induced drag is 50% of total drag at this speed out of ground effect - I really have no idea. Based on this wild assumption, a 23.5% reduction of induced drag is 1992 / 2 * 0.765 or 762 lbs, resulting in a total drag of 1758 lbs.
Plugging these numbers back in the climb gradient formula (1992 - 1758) / 14000 gives a climb gradient of 0.0167 or 1.6%. If this is correct, the aircraft can mush along just above a slightly rising runway whilst the undercarriage is retracted.
Am I on the right track? Is there anyway to improve the above calculations.