Centrifugal force problem

[thetexan]

I want to understand the problem of centrifugal force with an aircraft in a turn.

If I keep the wings level and input a certain amount of left rudder the aircraft skids and throws the turn coordinator ball to the right due to centrifugal force. The solution is to bank the wings until the ball becomes centered.

A similar situation occurs when you take a turn in a car on a level street. If you bank the road enough you will not feel the centfugal force as a side force but instead will experience a greater downward force, or "g"s.

My question is how do I calculate the amount of bank necessary to counter the side force in the car or in an aircraft for a given speed so that the is no apparent side force?

I know that in the case of the aircraft, to be coordinated in a turn, the horizontal component of lift must equal the centrifugal force. So how do I determine the centrifugal force for a given speed?

For example, a level coordinated turn with a bank of 60 degrees produces 2 gs. In a 4000 lb airplane the wings are supporting 8000 lbs, an increase of 4000 lbs. does that mean that there is 4000 lbs of centrifugal force in the opposite direction?

Again, what is the formula for this and how is it calculated? Thanks.

Tex

 

[haruspex]

If traveling at speed v and executing a turn of radius r, the centripetal force you need is mv²/r.
You also still need a vertical force of mg to counter gravity.
If banked at angle θ, your lift force, R, has to provide both:
R cos(θ)=mg
R sin(θ)=mv²/r.
Taking the ratio:
Tan(θ)=v²/(rg)

 

[sophiecentaur]

I want to understand the problem of centrifugal force with an aircraft in a turn.

If I keep the wings level and input a certain amount of left rudder the aircraft skids and throws the turn coordinator ball to the right due to centrifugal force. The solution is to bank the wings until the ball becomes centered.

A similar situation occurs when you take a turn in a car on a level street. If you bank the road enough you will not feel the centfugal force as a side force but instead will experience a greater downward force, or "g"s.

My question is how do I calculate the amount of bank necessary to counter the side force in the car or in an aircraft for a given speed so that the is no apparent side force?

I know that in the case of the aircraft, to be coordinated in a turn, the horizontal component of lift must equal the centrifugal force. So how do I determine the centrifugal force for a given speed?

For example, a level coordinated turn with a bank of 60 degrees produces 2 gs. In a 4000 lb airplane the wings are supporting 8000 lbs, an increase of 4000 lbs. does that mean that there is 4000 lbs of centrifugal force in the opposite direction?

Again, what is the formula for this and how is it calculated? Thanks.

Tex

To solve this one, you need to use the dreaded Vectors and Trig. The lift force, keeping the plane up, will always be equal to the weight (on a level flight) and the centripetal force will depend on what radius of turn you need for the particular speed. The wings will produce a force which, provides those necessary forces. See this link for some diagrams and explanations which include a banked turn in a plane. It shows you how the formulae that Haruspex provided (above) apply to the situation, with the aid of a diagram.

 

[A.T.]

I want to understand the problem of centrifugal force with an aircraft in a turn.

If the centrifugal force is a problem, then analyse the scenario in an inertial frame where the centrifugal force doesn't exist.

So how do I determine the centrifugal force for a given speed?

https://en.wikipedia.org/wiki/Centrifugal_force#Force