Hi Graham,
There is no need to guess. We can simply use normal BGS formulation to make a sound prediction.
V2/V1 = SQRT(W2/W1)
V1 = 300KEAS
V2 = ?
W1 = 10000lbs
W2 = 10500lbs
V2 = {SQRT(W2/W1)}*V1
V2 = {SQRT(W10500lbs/10000lbs)}*300KEAS
V2= 307.4 KEAS
If we hold angle of attack constant, we must increase our speed 7.4KEAS or 8.5 mph if we add 500lbs weight. Well one just might think that is insignificant. It certainly isn't gong to make much difference in what we can we catch or run from.
Now lets look at it from the Power required relationship in our fictional aircraft. Using standard BGS formulation for a power producer:
Pr1 = 2000thp
Pr2/Pr1 = (W2/W1)^3/2
Pr2 = {(W2/W1)^3/2}*Pr1
Pr2 = {(10500/10000)^3/2}*2000thp
Pr2 = 2680thp
Or a 34% increase in the amount of power required! That 4-5 KEAS in reality represents a very significant reduction in the designs power available.
Quote:
|
The induced drag, the only part affected by weight,
|
Not really. The induced drag is the only a direct affect of weight increase. The reduction in power available certainly does effect our ability to overcome the power of drag other than that due to lift. Remember the power of drag other than due to lift increases at the cube of velocity.
All the best,
Crumpp