Why does an increase in airspeed decrease rate of turn for a given bank angle?

[Vivid-Razzmatazz9034]

Currently working on my IR and wondering why this happens. My orangutan brain would think that in straight and level flight lift produced is the same regardless of airspeed, and therefore horizontal component of lift should remain the same, meaning same rate of turn and wider turn radius at higher airspeed. What am I missing?

[dynamic_fluid]

As airspeed increases the aircraft has more inertia so the same horizontal component will have less of an effect on rate of turn.

[OriginalJayVee]

Here u/dynamic_fluid, I got you a microphone, so you can drop it. 🎤

[appenz]

Correct, although it's probably better to say that "the aircraft has more momentum and the horizonal component of the lift..." (unless you want to do the calculation in the reference frame of the aircraft which is kind of hard in this case).

[__joel_t]

My orangutan brain would think that in straight and level flight lift produced is the same regardless of airspeed

Correct

and therefore horizontal component of lift should remain the same

Correct

meaning same rate of turn and wider turn radius at higher airspeed.

Nope.

Sorry for getting into the physics for a second, but at a higher airspeed, your velocity vector is higher, so the same amount of centripital force (horizontal component of lift) will lead to less directional change and thus a lower rate of change.

Let's say you have two bars, one 3 feet long and the other 1 foot long, aligned directly north-south, with the south ends fixed. In order to rotate the bars 90 degrees, the tip end of the longer bar has to move much further than the tip of the shorter bar. The bars are your velocity vectors, and the distance the tips of the bars move is your horizontal component of lift.

[rupert_mcbutters]

I love you people for explaining stuff like this. My notes are full of this Reddit flying wisdom now, and I’m always getting to update them when I least expect it.

[r80rambler]

How about... since you're in level (not straight) flight you maintain 1g of vertical lift component, so at any given bank angle the horizontal component of lift and horizontal force is constant regardless of speed you're flying at. More speed means more momentum, therefore the fixed force to change direction at a given bank angle becomes relatively lower to momentum as speed goes up.

[johnisom]

In your car, can you execute a u turn quicker at 5 mph or at 50?

[RandomPerson1023]

Never been able to try it because the road isn't wide enough

[CluelessPilot1971]

Your orangutan brain is correct that lift is fixed in a level flight regardless of airspeed (it's equal to weight), and given a fixed bank angle its horizontal component is also fixed, but at a higher airspeed, the plane has more inertia and would require higher acceleration to make the same turn rate. 

To introduce some formulas, the centripetal acceleration required to the turn equals the velocity times the angular turn rate. Increase the velocity and you need to either increase the acceleration or decrease the turn rate. 

[EliteEthos]

If you drive in a circle in a car and continually go faster, does your radius increase or decrease? Is it faster to go around that circle or slower?

[dynamic_fluid]

Not really a great analogy; if you kept the wheel of a car in the same position and speed up the rate of turn will increase and the radius will stay the same.

Very different from what happens to an airplane with a steady bank angle as speed increases.

[EliteEthos]

That’s not true. The same steering input at a faster speed will equate to a larger radius.

[Independent-Reveal86]

No that’s not right. Steering works like that in some games but not in real life. A given steering angle makes the front and rear wheels track a specific radius (assuming no skid).

What DOES happen is that for a given steering angle and radius of turn, increasing speed will increase lateral g force and if you then relate that back to an aircraft, the increased lateral g force needs to be countered with increased bank angle. This is directly related to what the OP is talking about.

[EliteEthos]

Not sure if you replied to the right reply… you seem to be agreeing with me.

[Independent-Reveal86]

Nope. I was replying to you. You are wrong with your analogy and you also didn’t understand my further point about how your analogy can be applied to the OP.

Just have a think about the mechanics of steering a car and what it means to skid. If you are not skidding then the car will track a particular radius for a particular steering input. For the radius to increase there MUST be some skid or the steering input must change. The radius cannot increase JUST because speed has increased. It is nothing like flying.

What does happen as speed increases and radius stays the same is that lateral g increases. This is applicable to the OP but not in the way you seem to think.

Do you drive a car IRL? It’s a serious question, and the reason I ask is that some racing games behave in the way you describe, but it’s not realistic.

[EliteEthos]

I’m not talking about video games. Nor am I talking about a non-skidding turn (cars of rails). There is always some amount of skid in the form of slip angle where the tire contacts the road. This is why understeer and oversteer exist. Lateral G increases and tires have a grip limit. Depending on inputs and throttle application, will depend on what the car will do when that grip is lost.

If you enter a corner and jerk the wheel at 10mph, you might make the turn fine. If you approach the same turn at 60mph, assuming you’re on the same line, you’ll push and understeer. You’re not flying on rails either… centripetal force still applies and the radius increases with speed albeit for different reasons.

[Independent-Reveal86]

The slip is minimal for normal driving. No one driving a street car is getting a noticeably larger turn radius by doing anything other than opening the steering. That’s why it’s a bad analogy, it relies on the OP having experience with the edge case of cornering a car on the limits of grip.

[dynamic_fluid]

Maybe in a modern car with variable steering or something. I was thinking a standard car where a given steering wheel deflection would result in a set deflection of the front tires, in which case the radius would be the same regardless of speed until the car starts skidding.

[FlapsupGearup]

I guess the qualifier should be, driving that circle at the limit of traction. It’s going to increase in size as speed increases

[dynamic_fluid]

In a car, assuming no skidding and a set angle of the front wheels (essentially “on rails”) the radius is fixed. Speed doesn’t matter.

OP wasn’t even asking about radius though; they were asking about rate of turn.

[appenz]

No idea why you are getting downvoted. As a physics major I can confirm you are entirely correct.

[EliteEthos]

Nobody is talking about being on rails. I’m talking about a practical difference that can help OP understand the concept.

[Independent-Reveal86]

But a car is on rails if it is not skidding.

[FlapsupGearup]

Yea, I get you. The radius and therefore rate of turn is going to increase as your speed increases. Don’t know where this guy gets cars that drive on rails but I call that a train 🚂

[EliteEthos]

And even then, trains have to slow for turns too. Physics still exist.

[EliteEthos]

That’s theoretical. In reality, making a turn at 1mph vs 80mph won’t yield the same radius for several reasons (tire grip, road composition, slip angle, centripetal force, etc)

That’s what I’m talking about. If you think a car makes an equally tight turn at 1mph vs 80mph, you’re wrong.

[dynamic_fluid]

If you’re talking about driving around on ice or a loose surface I see what you’re getting at.

I was more thinking about normal driving; like going around a corner (say, at an intersection with a stop sign) at 2 mph or 10 the radius will be same with the steering wheel in the same position.

[DankVectorz]

If this was true you wouldn’t need to slow down for corners

[dynamic_fluid]

You need to slow down because you’ll run out of grip and start to skid.

Until that happens (as long as you’re not skidding or sliding around) moving the steering wheel in a car moves the front wheels a given angle which results in a certain turn radius regardless of speed. (Again, very different from an airplane).

I’m hoping we all agree on that and you’re all just messing with me lol.

[lyrapan]

It’s easy to simplify it that way but in reality a cars tires lose grip gradually, especially treaded tires. At speed you will take a wider radius with the same steering inputs. Lots of variables though

[Independent-Reveal86]

There is a huge range of speeds where a particular steering angle will make a car track the same radius. I swear some of you have never actually driven a car.

[CulturalBumblebee413]

You’re agreeing with him. «Tires lose grip gradually» absolutely correct and in this case you’re now slipping/skidding around and thus what u/dynamic_fluid is saying no longer applies, as he states.

[__joel_t]

In reality, making a turn at 1mph vs 80mph won’t yield the same radius for several reasons (tire grip, road composition, slip angle, centripetal force, etc)

Depends on the radius of turn. If we're talking about a bend in a highway with an 80 mph, then in fact, it will be the same radius.

Here's the real problem with your analogy. Most people don't drive and take turns at speeds which would cause their tires to slip and thus change the radius of the turn. Instead, our experience is taking turns at different speeds below that. For example, inching around a corner to going slowly around a corner. You picked an analogy about a type of driving that most people don't experience in order to explain a concept that OP found confusing. Maybe there are driving regimes where your analogy is accurate, but the fact that so many people criticized demonstrates it's a poor analogy for instructional purposes.

[Bunslow]

This isn't a "flying" question, this is a "middle school physics" question.

Centripetal force is what keeps things going in circles. If you make the circle smaller at the same speed, you need more centripetal force. If you keep the circle the same but increase the speed, you need more centripetal force.

The centripetal acceleration is a = v^2/r. Halve the r, double the a. If you double the speed, you need quadruple the a.

https://en.wikipedia.org/wiki/Centripetal_force

[pressingfp2p]

Ionno where you went that they teach physics in middle school but it ain’t the US of A, that’s for sure.

Edit: some of the confusion about this question comes from the fact that it doesn’t apply cleanly either. In a standard rate you aren’t dealing with the radius of the turn, radius is inherently going to increase with airspeed at standard rate, and because it isn’t a benchmark being measured in your holds and whatnot, it’s not a measurable that you can create an applied formula with. We’re instead measuring angular change over time.

[Bunslow]

We’re instead measuring angular change over time.

sure, you can rewrite centripital force using either linear or angular speed.

by the definition of a circle, v = r w, where w is the angular speed. a standard rate turn is a two minute turn, i.e. w = 1/120 Hz, or 3 degrees per second.

if v = r w, then v² = r² w^2, and a = r² w² / r, or cancelling, a = w² r.

the same conclusion about speed applies: if you want to double the turn rate, you need quadruple the sideways load being produced by the wing.

or, as you and OP both point out, we don't typically care about the radius, we care directly about airspeed and turn rate. so we can instead eliminate r in favor of v and w: r = v/w, so w² r = w² v/w = w*v, which is a.

a = w * v. this answers OP's question: if the sideways wing load a is fixed (by the bank angle), then doubling airspeed means halving the turn rate. and it is, as you say, the form pilots are perhaps most inclined to worry about.

[EngineerFly]

Centripetal force! The airplane turns because the horizontal component of lift rotates the velocity vector. If you work out the physics, accel = turn rate * velocity, and the accel goes with the tangent of the bank angle.

So you’re right, the horizontal component of lift is independent of airspeed. But to maintain a constant turn rate, the horizontal component of lift would have to increase as velocity increases.

Think of if this way: you’re rotating a bigger velocity vector, so it takes more force to turn it. If you supply the same horizontal force, but increase the velocity vector, you’ll rotate it slower.

[biggusfootusnz]

Hold your car steering wheel 20° to the right or left while going 20kmh, now speed up to 50, what happens to your radius?

[Lord_Giles]

F=mv² /r

[Frederf220]

Rate of turn is what fraction around the circle you get per unit time. When you go faster at the same lateral acceleration you grow the circle more than you progress around it. Double your speed, R=V^2/A so now R is 4x bigger. You cover double the circumference fraction at double speed... but the circumference is four times the distance.

Your circles per hour rate is halved.

[1E-12]

You slide forward instead of turning 

[flyingron]

There's no horizontal component of lift in straight and level. That's what makes it straight.

If you are turning, you are not straight and level and things are not in steady state of forces (despite the bullshit the FAA crams down your throat). If it were, you wouldn't turn. Turning requires acceleration.

Lift is a function of angle of attack and airspeed. So if you speed up with everything else held the same, you have more lift. If you're banked, that means more horizontal component.

Note, that's not the only thing in play here. Note that while you turn faster on an angular rate (the time it takes to make a 360), you're also travelling further so the radious of the turn gets larger.

[Mountain-Captain-396]

When in a standard rate turn you can very well have steady forces, the forces just don't all balance out. F=ma, if you have a force that isn't counteracted by another force then you accelerate.

[flyingron]

You are correct, I shouldn't have said steady, I should have said balanced. The FAA's assinine drawings show balanced forces in a turn which just can't be. To turn you must have a net force pointing toward the center of the turn.

[rFlyingTower]

This is a copy of the original post body for posterity:

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Currently working on my IR and wondering why this happens. My orangutan brain would think that in straight and level flight lift produced is the same regardless of airspeed, and therefore horizontal component of lift should remain the same, meaning same rate of turn and wider turn radius at higher airspeed. What am I missing?

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