Is the ratio not dependent on the aircraft thrust to mass ratio?
Peter wrote:
Is the ratio not dependent on the aircraft thrust to mass ratio?
Yes to an extent in that the higher the ratio of thrust to weight, the less a fixed amount of gravity drag will affect you. But a rule of thumb is usually good enough. If you need to calculate that exactly you should prob reconsider the departure.
JasonC wrote:
Of course not but where would we be seriously talking 20%? It is a rule of thumb for a normal runway where you derive it by the drag component of the gravity vector.
Peter wrote:
Is the ratio not dependent on the aircraft thrust to mass ratio?
Yes on small angles distance penalty is proportional to loss of forward acceleration which is 1g per 1% slope gradient, hence the +10% (basically +g=9.81%), ignoring lift-to-drag ratio and friction coefficient, when slope gradient exceeds aircraft thrust-to-weight ratio there is no forward acceleration, but that rule of thumb “10% for 1%” should work when slope % is way less than thrust-to-weight ratio
Forward acceleration on wheels: [(thrust-to-weight) – (slope gradient) – (friction ratio)] + [(lift-to-weight)(friction ratio)(1-(drag-to-lift ratio))]
Forward acceleration on wings: [(thrust-to-weight) – (climb gradient) – (drag-to-lift ratio)]
Lift-to-weight does not matter at slow speeds, so takeoff roll distance is mostly a question of thrust-to-weight at slow speeds, slope gradient, friction ratio and liftoff ground speed (e.g. stall-wind)
Ibra wrote:
Yes on small angles distance penalty is proportional to loss of forward acceleration which is 1g per 1% slope gradient, hence the +10% (basically +g=9.81%),
The gravity drag is not 1g per 1%! How on earth did you get that?
I’m certainly not gonna calculate any of that before I go flying.
We have a slight slope at base.
We regularly get 90 deg xwinds.
Heavy A/C & crossword, downslope always.
Light A/C not so important.
With 5kts tailwind…..won’t take many meters to get the extra 5 kts on the ground while heading downslope.
Calculations like that, margins of MAUW against length of grass/rwy length/softness of surface etc, is for ferry pilots and test pilots.
If I need to consider getting a calculator it’s too tight.
JasonC wrote:
The gravity drag is not 1g per 1%! How on earth did you get that?
No, 1g per 1% is the 10% increase in takeoff distance per slope gradient % !
Ok let’s teach some approximate physics,
- If you assume small angles, then gravity drag is -Weight*Sin(Angle) which is mass*g*gradient% for small angles (Sin = Tan = Angle in Radian), does this look correct?
- If you assume constant acceleration, then Acceleration*Distance = Speed^2 = (StallSpeed-Wind)^2 on flat and on gradient, does this look correct?
So (FlatAcceleration-g*gradient%)*Distance(gradient%) stays the same, this formula should give about 10% increase of distance for each 1% of slope gradient, obviously this does not work when gradient % is very high (takeoff distance shoot through the roof when gradient% = FlatAcceleration/g = thrust-to-weight)
It is most of the time better to land uphill and take-off down hill even with a slight tailwind. As Jason already mentioned, a simple rule of thumb can be used to calculate the runway gradient: every 1.0º grade equals approximately 10% change in effective runway length.
Some rules of thumb that can be used:
Down-slope take-off distance is reduced by about 5% per degree of slope.
Up-slope take-off distance is increased by about 7% per degree of slope.
Is it degrees or percent like Jason wrote? Because that makes a big difference.
@Clipperstorch: you are right. I meant %.
Ibra wrote:
- If you assume small angles, then gravity drag is -Weight*Sin(Angle) which is mass*g*gradient% for small angles (Sin = Tan = Angle in Radian), does this look correct?
I’m not entirely following this are you making a direct substitution for gradient percentage to radians without converting? a 100% gradient would only be pi/4 rad.