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.
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asked Aug 9 '17 at 11:56
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$\begingroup$ 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? $\endgroup$ – Penguin Aug 9 '17 at 12:24
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$\begingroup$ It depends a lot on the weight. Thrust is independent on weight, but break force is almost linear to weight. $\endgroup$ – user3528438 Aug 9 '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.
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$\begingroup$ That's consistent, but considerably lower than I expected. $\endgroup$ – Daniele Procida Aug 9 '17 at 15:02
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$\begingroup$ Yes, I remember learning a long time ago that hard braking was actually close to 1g but could not find a reference to it. $\endgroup$ – Koyovis Aug 9 '17 at 15:07
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$\begingroup$ Does it mean that at a dry concrete, a plane with TO thrust and full brakes won't ever move? $\endgroup$ – yo' Aug 9 '17 at 15:07
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$\begingroup$ @yo' It also needs to be at TO load. The ratio could easily reverse if the plane is not heavily loaded. $\endgroup$ – user3528438 Aug 9 '17 at 15:53
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$\begingroup$ @user3528438 Ah right! The max friction force is of course proportional to weight! Thanks for the hint! $\endgroup$ – yo' Aug 9 '17 at 15:59
