An aerodynamics question...I think?

[mscard88]

Nah, I’m thinking of the times that the nose may fall through the horizon so you add back pressure while you’re in the turn. Isn’t that negative?

That be positive my friend.

 

[RyanB]

Nope, that's just garden-variety "positive" g's. The kind that make you "feel heavier".

Negative g's would make you "feel lighter". Like if you do a "zoom over the top maneuver" and everything in the plane starts floating.

Ah, so I had my g’s confused.

 

[Salty]

I must not be up to speed with my various g forces. What’s the term for the g felt during a steep turn?

Just because you don’t like them doesn’t make them negative.

Thems positive g’s you’re talking about.

 

[RoscoeT]

If you don't like positive G, then I guarantee you're really gonna hate negative G.

 

[Salty]

Negative g’s are the butterflies in the stomach or standing on your head feeling. Positive g’s are the “my face is melting” type.

 

[kath]

The normal force doesn't have to be a fraction of the car's weight, though for you and me it always is...

All true... edited in original post. (<gulp> My physics cred is at stake!)

 

[RoscoeT]

Negative g’s are the butterflies in the stomach or standing on your head feeling. Positive g’s are the “my face is melting” type.

Zero G is "butterflies". Initial impression of real negative G (-3G and more) is more like OMG my skull is going to explode please make it stop...then you get used to it.

 

[genna]

Not if the road is flat. Friction between tires and a road will max out at some fraction of the vehicle's weight, because the force = (mu)*N, where "mu" is a number generally between zero and one (like 0.2 or something), and "N" is the upward force necessary to hold the car up**.

Airplane wings, on the other hand, will max out at something like 3.8 times the vehicle's weight (if flying within normal category!). More, if utility/acrobatic. Even at a modest 45-degree bank, where the horizontal component is only 1/sqrt(2) of this, this is better than the car.

The normal force from a banked piece of pavement, on the other other hand, will max out at some astoundingly huge multiplier of the vehicle's weight.

So in the hierarchy of turnability (assuming the speed is fixed at 100 mph or whatever): wheels on the flat road is the worst performer, airplane does better, and banked road can do the tightest turns assuming it's banked steep enough (like a luge track!)


Now, boats, on the other other other hand, are going to be complicated...

<Physics, yay!>

** Edit: ..."N" is equal to the car's weight, for most everyday cars.

Except the 3.8 you are talking about also supports the plane's weight, while in the car you are only talking about lateral force. Additionally, wings (race cars) can provide quite a lot of lateral traction. F1 cars can have close to 8g lateral acceleration at high speeds

 

[kath]

Additionally, wings (race cars) can provide quite a lot of lateral traction. F1 cars can have close to 8g lateral acceleration at high speeds

Not knowing much about racing, I imagine racecar tires are also much "grippier" than on a normal car, with a "mu" coefficient that is much higher. This, together with the larger "N" from the downforce, will allow a centripetal force that can give the airplane a run for its money.

<Materials science, yay!>

 

[Shepherd]

All aircraft skid in a turn. You experience it as "G" force in a coordinated turn. The plane will change direction, but will resist the change and try to continue in the original direction until all the mass of the plane gets with the program.
Vector analysis can show you the deviation from your course.

An object in motion, etc


Car tires were so crummy, that when I was drag racing it was believed that no car could accelerate faster then 1G.
The Big Daddy Don Garlits stood the racing world on it's ear. Soon, multiple "Gs" in a turn was the norm as tire manufacturers started experimenting with new compounds.

 

[genna]

Not knowing much about racing, I imagine racecar tires are also much "grippier" than on a normal car, with a "mu" coefficient that is much higher. This, together with the larger "N" from the downforce, will allow a centripetal force that can give the airplane a run for its money.

<Materials science, yay!>

F1 is , of course, an extreme example with downforce far exceeding the weight of the car, but it’s pretty easy to go over 1g turning in lightly modified street car even on street tires. Just need some downforce

Also, you can go well over tire limitations by performing Scandinavian flick

 

[wayne]

Driving a car on the road at high speeds requires one to slow down in order to make a turn. In other words, if you’re traveling at 100mph on a straight road and then a 90deg turn to left is ahead, you would have to greatly reduce speed in order to make that sharp of a turn without going off the road and doing extreme damage to the vehicle.

Or bank the road more. Just like NASCAR.

 

[Bill]

F1 is , of course, an extreme example with downforce far exceeding the weight of the car, but it’s pretty easy to go over 1g turning in lightly modified street car even on street tires. Just need some downforce

Also, you can go well over tire limitations by performing Scandinavian flick

Yup, the flick loads up the outside tires, N = more g's.

And bikes routinely corner well above 1g, a motorcycle at a 45* lean is 1.4g's, take that to 50*, and it's 1.67g's. Of course when I ride like that I'm lucky to get 3500mi out of a set of tires.

 

[genna]

Yup, the flick loads up the outside tires, N = more g's.

And bikes routinely corner well above 1g, a motorcycle at a 45* lean is 1.4g's, take that to 50*, and it's 1.67g's. Of course when I ride like that I'm lucky to get 3500mi out of a set of tires.

 

[RyanB]

If you are asking why cars need to slow down and planes don't, the answer is the radius of the turn. If your 90 turn on the road was over 5 miles, you would not need to slow down in the car either.

Airliners take many miles to make a turn for that exact reason. Not to stress the plane and occupants

It's just hard for my brain to grasp the speed and abilities of jet aircraft at times. Of course I realize that airplanes and road vehicles are not bound to the same laws of physics, but good point nonetheless.

 

[genna]

It's just hard for my brain to grasp the speed and abilities of jet aircraft at times. Of course I realize that airplanes and road vehicles are not bound to the same laws of physics, but good point nonetheless.

This is where your mistake is. There is only one set of laws of physics(Quantum physics not withstanding). Planes and cars are bound by the same laws. Your scenarios are so vastly different, however, that comparing them makes very little sense. If you bring car to the speed and the radius of the turn of an airliner, you will see that it behaves remarkably similar. You will not need to slow down and you will hardly feel any Gs. The main difference is how the lateral acceleration is achieved. Car uses tire friction, plane uses portion of the wing's lift.

 

[RyanB]

This is where your mistake is. There is only one set of laws of physics(Quantum physics not withstanding). Planes are cars are bound by the same laws. Your scenarios are so vastly different, however, that comparing them makes very little sense. If you bring car to the speed and the radius of the turn of an airliner, you will see that it behaves remarkably similar. You will not need to slow down and you will hardly feel any Gs. The main difference is how the lateral acceleration is achieved. Car users tire friction, plane users portion of the wing's lift.

that's why I said I wasn't a very good physics student. You know what I'm saying...

 

[genna]

Yup, the flick loads up the outside tires, N = more g's.

And bikes routinely corner well above 1g, a motorcycle at a 45* lean is 1.4g's, take that to 50*, and it's 1.67g's. Of course when I ride like that I'm lucky to get 3500mi out of a set of tires.

That's not the turning G, however. That's total G. At that 68* degree that Marquez manages to do in GP, the side acceleration is only ... 2.47G(total G is 2.66)

Nevermind it's still pretty high

 

[luvflyin]

Ah, so I had my g’s confused.

That's what she said

 

[luvflyin]

I was looking for those ice racers with spikes stickin out of their tires. Found this. Where was she back in.......

 

[Bill]

Marquez is unholy! How he does the things he does...

 

[genna]

Marquez is unholy! How he does the things he does...

Indeed! I don't have a clue. But consider that the 68* angle is actually more than that. The angle shown is the motorcycle lean angle, not the CG lean angle(that which matters for turning) which is higher by a few degrees due to body position.

 

[tspear]

So how does the bike exceed 1 G of lateral force?

Tim

 

[genna]

So how does the bike exceed 1 G of lateral force?

Tim

the weight(vertical component of the force) is 1G always. At the lean angle of 68*, the horizontal component of the force is 1G*tan(68)=2.47G and total force experienced by the rider is 1G/cos(68)=2.66G.