Menu Sign In Contact FAQ
Banner
Welcome to our forums

Depart uphill or downhill?

Is the ratio not dependent on the aircraft thrust to mass ratio?

Administrator
Shoreham EGKA, United Kingdom

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.

EGTK Oxford

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)

Paris/Essex, France/UK, United Kingdom

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?

EGTK Oxford

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.

United Kingdom

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)

Last Edited by Ibra at 20 Jun 23:35
Paris/Essex, France/UK, United Kingdom

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.

EDLE, Netherlands

Is it degrees or percent like Jason wrote? Because that makes a big difference.

EDQH, Germany

@Clipperstorch: you are right. I meant %.

EDLE, Netherlands

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.

Sign in to add your message

Back to Top