Test Pilot time: Short field landings.

[tspear]

I wish I had spent more time and effort in physics and math classes. With that said, the braking force, which comes frictional forces opposing forward motion, is a function of the force applied normal or perpendicular to the runway.

In the case of an aircraft, that force maxes out at the weight of the aircraft which I’m thinking is equivalent to 1G.

In the case of automobiles, particularly race cars, the force can be more than 1G. It can be the weight of the car plus aerodynamic down force.

With aircraft, we struggle to get all the weight born on braking wheels; 2 point rollout, spoilers, retracted flaps, minimal angle of attack.

With automobiles all the weight is already on all 4 braking wheels, there is no lifting force, and in the case of race cars, there can be significant downforce. Downforce increases friction between the tires and road surface which can be used to keep the car on the road and for increased braking force.

Am I getting close to something?


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You have it. Race cars are known to get as much as 5g of downforce. Airplanes actually have less than one G as the wings provide lift reducing the available down force.
Technically with adhesion and other factors you may be able to exceed one G by a minimal amount.

Tim ( not an engineer)

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[SoCal RV Flyer]

You have it. Race cars are known to get as much as 5g of downforce. Airplanes actually have less than one G as the wings provide lift reducing the available down force.
Technically with adhesion and other factors you may be able to exceed one G by a minimal amount.

Tim ( not an engineer)

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Even without aero downforce, it's possible to generate roughly 1.3g of lateral and braking acceleration. Go-karts are a good example. It's a function of the mechanical interaction between a soft tread and the minor imperfections in the track surface...think of meshing gears on a microscopic level.

 

[tspear]

Even without aero downforce, it's possible to generate roughly 1.3g of lateral and braking acceleration. Go-karts are a good example. It's a function of the mechanical interaction between a soft tread and the minor imperfections in the track surface...think of meshing gears on a microscopic level.

Interesting on the 1.3G. I do not recall it being that high.

Tim

 

[SoCal RV Flyer]

Interesting on the 1.3G. I do not recall it being that high.

It's somewhat surface-dependent. I was involved in some skidpad testing with the karts, and it was on a textured concrete surface. High tire wear but great grip!

 

[flyingcheesehead]

I'm not an engineer and my knowledge of this one-G maximum braking phenomenon came from an ex-National Guard pilot I used to fly with who was familiar with aircraft brake testing experiments done at Edwards AFB long ago. He said to imagine a pendulum of one "G". As you brake, it swings forward, but can't go higher than horizontal, at which point there would be no more "G" force on the runway. I accepted the principle, but always suspected the explanation. Maybe one of the resident engineers here can enlighten me/us?

1 G deceleration would put the pendulum at 45º, not horizontal.*


* Yes, for those who nitpick like I sometimes do, we have to assume an idealized "pendulum" that didn't swing, otherwise we'd have to have a low-jerk brake application. And yes, I mean Jerk the technical term - The derivative of acceleration.

 

[dtuuri]

1 G deceleration would put the pendulum at 45º, not horizontal.

That makes sense, but from where does the 1 G stopping limit, whether just practical or mathematical, come from then? I would guess that tire tread is designed to minimize friction during the takeoff roll, does that have something to do with it?

 

[flyingcheesehead]

That makes sense, but from where does the 1 G stopping limit, whether just practical or mathematical, come from then? I would guess that tire tread is designed to minimize friction during the takeoff roll, does that have something to do with it?

There isn't a 1G stopping limit. Intuitively, you'd think there would be, but it depends on a lot of things - namely, anything that change friction.

Tire tread isn't designed to "minimize friction during the takeoff roll" - You don't want your tires to skid during takeoff either.

I would imagine that the reason we probably don't see more than 1G very much is that you wouldn't want aircraft tires to have too soft a tread, as normal landings would flat-spot them if that were the case. But, a harder tread won't "bite" the pavement as well.

 

[kath]

That makes sense, but from where does the 1 G stopping limit, whether just practical or mathematical, come from then?

The maximum the force of static friction (that is, "grip" without slipping) can be is "mu * N". (Well, reality is more complicated than this, but this is a freshman-physics-approximation kind of treatment.)
"N" is the normal force from the pavement. If the surface is flat and there's no racecar-like downforces or lift forces or anything like that, this is the weight of the aircraft.
"mu" is a coefficient which is typically between 0 and 1.
So if "mu = 1.0", then the sideways force of friction/"grip" can be (at most) equal to the downward weight of the aircraft.

Now, there's no law of physics that says "mu" has to be 1.0 or less. I can be more than 1.0, thanks to the wonders of materials science. But mu's less than 1.0 are typical for most everyday household surfaces-gripping-against-other-surfaces.

 

[PaulS]

Pop those flaps out if you need to stop quickly in the planes I fly. Brake hard with them down and you will hear a lot of squealing, especially from an instructor if one is on board.

 

[dtuuri]

The maximum the force of static friction (that is, "grip" without slipping) can be is "mu * N". (Well, reality is more complicated than this, but this is a freshman-physics-approximation kind of treatment.)
"N" is the normal force from the pavement. If the surface is flat and there's no racecar-like downforces or lift forces or anything like that, this is the weight of the aircraft.
"mu" is a coefficient which is typically between 0 and 1.
So if "mu = 1.0", then the sideways force of friction/"grip" can be (at most) equal to the downward weight of the aircraft.

Now, there's no law of physics that says "mu" has to be 1.0 or less. I can be more than 1.0, thanks to the wonders of materials science. But mu's less than 1.0 are typical for most everyday household surfaces-gripping-against-other-surfaces.

Thank you for this. Nice explanation.

 

[FastEddieB]

It always seemed to me that once you’re generating more than 1g, you should theoretically be able to get the car to stick to a wall.

Kinda like a rotating space station could generate 1g so astronauts could jog around the inside.

No?

 

[flyingcheesehead]

It always seemed to me that once you’re generating more than 1g, you should theoretically be able to get the car to stick to a wall.

Kinda like a rotating space station could generate 1g so astronauts could jog around the inside.

No?

No... Depending on what you mean here.

As @kath mentioned, one of the primary parts of the friction equation is the normal force. If you just plopped a car against a vertical wall, the normal force would be zero (since gravity would now be a tangent force and not a normal force), so the car comes crashing back down to the ground.

But, using your "rotating space station" example, if you had a larger version of the amusement park ride with the big cylinder that spins around with people around the inside and the floor drops away (sorry, can't remember what it's called), it would work. Basically, if you had a "wall" that was the inside of a cylinder, put the car up against the inner part of the wall and rotated it around at a speed that resulted in a 1G force in a downward direction relative to the car (away from the center of the cylinder), then yes, you absolutely could stick a car that's able to accelerate at >=1G up against the wall and it'll "stick".

 

[MauleSkinner]

The maximum the force of static friction (that is, "grip" without slipping) can be is "mu * N". (Well, reality is more complicated than this, but this is a freshman-physics-approximation kind of treatment.)
"N" is the normal force from the pavement. If the surface is flat and there's no racecar-like downforces or lift forces or anything like that, this is the weight of the aircraft.
"mu" is a coefficient which is typically between 0 and 1.
So if "mu = 1.0", then the sideways force of friction/"grip" can be (at most) equal to the downward weight of the aircraft.

Now, there's no law of physics that says "mu" has to be 1.0 or less. I can be more than 1.0, thanks to the wonders of materials science. But mu's less than 1.0 are typical for most everyday household surfaces-gripping-against-other-surfaces.

My freshman physics approximation was a statistical failure. So was my final grade.