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Is the ratio fairly constant across aircraft types, sizes and powerplants?

I'd like to ignore the reverse thrust component, since that's not available at the same time as forward thrust.

Daniele Procida
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  • Can you clarify what you mean by "thrust braking force ratio" if it's not the ratio of max forward thrust to max reverse thrust? – Penguin Aug 09 '17 at 12:24
  • It depends a lot on the weight. Thrust is independent on weight, but break force is almost linear to weight. – user3528438 Aug 09 '17 at 15:50

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If we assume that the braking system is designed to leave no unused braking capacity, the maximum braking deceleration is determined by the friction coefficient of the tyres. This NASA report gives a friction coefficient of around 0.5, for dry concrete with a functioning anti-skid system. This results in maximum braking deceleration of 0.5 g, around 5 $m/s^2$

A typical maximum thrust acceleration would occur for a typical airliner, at TO thrust, and medium range TO weight. Let's take typical numbers for an A320:

  • A320: TO thrust = 2 * 120 = 240 kN, TO weight = 70,000 kg, a = F/m = 240/70 = 3.5 $m/s^2$
  • B777: TO thrust = 2 * 500 = 1000 kN, weight = 300,000 kg, a = 1000/300 = 3.3 $m/s^2$
  • Fokker 100: TO thrust = 2 * 67 = 134 kN, weight = 40,000 kg, a = 134/40 = 3.35 $m/s^2$

So across the board, the braking-to-thrust force ratio is around 5/3.3 = 1.5.

Koyovis
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  • That's consistent, but considerably lower than I expected. – Daniele Procida Aug 09 '17 at 15:02
  • Yes, I remember learning a long time ago that hard braking was actually close to 1g but could not find a reference to it. – Koyovis Aug 09 '17 at 15:07
  • Does it mean that at a dry concrete, a plane with TO thrust and full brakes won't ever move? – yo' Aug 09 '17 at 15:07
  • @yo' It also needs to be at TO load. The ratio could easily reverse if the plane is not heavily loaded. – user3528438 Aug 09 '17 at 15:53
  • @user3528438 Ah right! The max friction force is of course proportional to weight! Thanks for the hint! – yo' Aug 09 '17 at 15:59
  • Information indicates that to calculate a disk braking force, you also need to know the coefficient of friction of the brake pads, the force the pads apply, how many there are per wheel, and the radius at which they act on the disk; for each wheel. The size of the tyre contact patch I expect should also be a factor. This make engineering sense that these factors all effect brake performance. What do you think, or did I misunderstand something? One might be able use v^2 = u^2 + 2as for the landing roll, but this assume linear deceleration, and the plane comes to a complete stop. – Penguin Aug 10 '17 at 03:55
  • @Richard we're doing an estimate here, not unlike what is done in the pre-design phase of an aircraft. What we don't want to do at this stage is lose ourselves into details before we've gained a ROM number. Looking at this from a helicopter view, it makes sense if the brake system is designed such that the friction between wheel and runway is the limiting factor, otherwise there is brake capacity left that cannot be effected. – Koyovis Aug 10 '17 at 05:41