What landing distance

Airborne_Again wrote:

For one thing speed is not constant but reducing into the flare, also when you make a normal (as opposed to deadstick) landing, you will use some power all the way to the flare

I wouldn’t think too much about speed but rather about energy (the lighter plane will be slower or at most as fast as the heavier one at any given point in time).
In my model I only look at energy: Both planes start at a certain Energy E_1 that is basically their individual mass times the (equal) approach speed squared. At a lower Energy E_S the glide part of the approach ends and that is basically their mass times the (individual) stall speed squared – and it is the role of the pilot to make sure that this energy is reached exactly when the wheels touch ground.
From that energy level the plane is rolling on the runway and we brake until the energy is 0.

For that second part, -dE/dt is a function of the maximum brake power. That maximum brake power scales lower than linear with the weight of the plane (as static friction is lower than linear with normal force) leading to the well experienced result, that when braking a car with same speed the heavier one has the longer braking distance.
It’s worth noting that the touchdown speed and therefore the ground roll in this model is independent of the initial approach speed and therefore ground roll book values for the individual weights hold true.

For the first part one could see in the Archer example, that while E_s is substantially higher for the heavier plane, the difference between E_1 and E_s is still smaller for the lighter one.
The core question (even with power) would therefore be: Is there a reason, why -dE/dt should be significantly higher for the heavier plane than for the lighter plane during the glide phase of the approach? At the same power setting I would even assume it is lower, as the heavier plane converts more potential energy per time (at same sink rate).

Malibuflyer wrote:

but that it only relevant for the part of the deceleration after touchdown.

But this basically is your landing distance? otherwise you could have more effective brakes. Drive a car sportingly and one quickly learns how important weight on the tyres is for grip

There’s also the fact that many landing surfaces have inclines, which will have a significant effect.

Practically this probably comes down to individual aircraft and the landing site in question.

I understand for aircrafts operated with an AoA you have an “approach/flare AoA targets” if you fly those values on any approach approach from “approach AoA” to land with “max AoA”, then your flare & ground rolls should be independent of aircraft mass, the same applies to ground friction as mass inertia just cancels out…

Given the difficulty in flying on AoA when people will come to land, some will just figure out the associated speed (depends on mass & config) and fly that speed on approach for their given weight & config

So flare roll & ground roll are roughly invariant vs weight if you think in terms of “target AoA” and max breaking, but people will tend to fly different AoA values as they tend to fly different speeds & weights depending on configs & conditions and also apply breaks differently…

The invariance vs weight does not hold for gusts/updrafts components or under ground friction hysteresis components,
- For aerodynamics, weight matters a lot for RoC, RoD and decelerations as AoA oscillates (heavy aircraft don’t lose energy quickly when hit by gusts)
- For ground friction, weight matters a lot for sliding hysteresis when rubber start to make sounds (don’t know the phenomenon name in English but it is “fixe-gliss regimes” in French)

Off_Field wrote:

Drive a car sportingly and one quickly learns how important weight on the tyres is for grip

Force on tyre is important for grip – not weight! Therefore sport cars are built as light as possible but have an aerodynamic design that creates much downforce.

If you just add another 100kg weight to any car, all of its relevant performance figures (acceleration, braking distance, max turning g forces, etc.) will go down with exception of very special situations at the fringe of it’s performance envelope (e.g. if friction on snow is so low that acceleration is zero, adding weight could raise acceleration above zero).

Ibra wrote:

I understand for aircrafts operated with an AoA you have an “approach/flare AoA targets” if you fly those values on any approach approach from “approach AoA”

Agree – but if you fly a constant “approach AoA” (which is exactly right in practice) your approach speed in a lighter plane will “automatically” be lower. Hence that does not contribute to the answer to the original question “how is landing distance affected if I don to reduce my approach speed according to the below MTOW weight?”

Ibra wrote:

So flare roll & ground roll are roughly invariant vs weight if you think in terms of “target AoA” and max breaking,

Not really: At “target AoA for touchdown” you will be much faster in a heavy plane than in a lighter one. Therefore ground roll is substantially longer for higher weight (as braking distance at same speed is already longer with a heavier vehicle and in addition you have more speed).

Malibuflyer wrote:

Force on tyre is important for grip – not weight!

Weight is a force.

Malibuflyer wrote:

as my own plane has no weight vs. stall speed diagram in the POH

This is a function of the square root of the actual weight to the MAUW. So a 2300 lb MAUW aircraft with a stall of 50 KCAS, will stall at 2000 lbs at 1 G at sqrt 2000/2300 times 50 (around .94) for an adjusted stall speed of 47 KCAS. Note the use of KCAS and not KIAS, correct short field technique will use the KCAS calibration chart to derive the correct KIAS.

As a rule of thumb the new LDR should work out at around twice the % reduction in stall speed for the reduced weight. YMMV

Malibuflyer wrote:

Not really: At “target AoA for touchdown” you will be much faster in a heavy plane than in a lighter one…Hence that does not contribute to the answer to the original question “how is landing distance affected if I don to reduce my approach speed according to the below MTOW weight?”

Yes I agree, the only “residual dependency on mass” comes from “touchdown speed” and “breaking strategy”, the former relates more to the +10kts winds and pilot technique (selecting approach speed) than 3kts change in stall speeds from weight , the latter is just random to quantify, so not sure where that cut through,
- If aerodynamic breaking (no toe breaks) first, you are almost in same “aerodynamic story” as flare as in ground roll when the weight is still on the wings and not entirely on the wheels at higher speeds
- If toe breaks, the answer is purely kinetic energy and heat dissipation rate to get a distance, this grows quadratic in mass if you apply them exactly on touchdown (from mass and from stall speed mass), linearly in mass if you apply breaks at specific ground speed value rather than touchdown and for people who wait for 10kts GS to apply foot breaks, the mass impact is will be really small if approach speed is dead where it should be….

This chart on accelerate – stop is interesting. Because an MEP has a take off safety speed which is in part a function of Vmc, and independent of mass (Vmc actually increases at lighter mass/weight due to the horizontal component of lift element in the allowance of up to 5 degree bank into the live engine), ASDR increases as weight decreases, which is counterintuitive.

If Vtoss in an MEP was just a function of mass, then ASDR would reduce with mass, as the lift off speed reduced. But here we have a speed which at lighter mass implies little or no weight on the wheels, and less braking action, therefore ASDR increases in inverse proportion to mass.

RobertL18C wrote:

This chart on accelerate

What aircraft is this for? I just checked a few poh, in my library, and the ASDA decreases with weight.