Banked turn

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A banked turn is the term used to describe a car riding along a circle with inclined edges. The angle at which a turn is banked refers to the angle of incline of the given path. The benefit of such a structure is that there are forces other than that of friction to keep the car on its designated path. Banked turns also have applications to aviation.

Contents

[edit] Flat surfaces

To begin our analysis of banked turns we shall first consider the forces acting on a car traveling on a flat circle. In this situation, the force that is keeping the car on its path is that of friction. The goal of our analysis will be to determine a minimum/maximum speed a car can travel to remain on its path. An important note is which way friction points in our analysis. When considering maximum velocity, friction points towards the center of the circle, as the car is being "pushed" outside and friction opposes the direction of motion. For minimum velocity the opposite is true. For the purpose of this example we will only analyze maximum velocity. Because friction in this case acts as the centripetal force, or the force "pulling" the car into the circle, we have the following equation:

{mv^2\over r} = \mu mg

The left side of the equation is the equation for centripetal force while the right side is friction, or the coefficient of static friction multiplied by the car's normal force. Solving this equation for velocity we get:

v = {\sqrt{r\mu g}}

[edit] Frictionless banked turns

As opposed to a car riding along a flat circle, inclined edges add an additional force that keeps the car in its path and prevents it from being "dragged into" or "pushed out of" the circle. This force is the horizontal component of the car's normal force. In the absence of friction, the normal force is the only one acting on the car in the direction of the center of the circle. Therefore, we can set centripetal force equal to the horizontal component of the normal force:

N\sin \theta ={mv^2\over r}

Because there is no motion in the vertical direction, the sum of all vertical forces acting on the system must be zero. Therefore we can set the vertical component of the car's normal force equal to its weight:

Ncosθ = mg

Solving the above equation for the normal force and substituting this value into our previous equation, we get:

{mv^2\over r}= {mg\tan \theta}

Solving for velocity we have:

v= {\sqrt{rg\tan \theta}}

This provides the velocity that in the absence of friction and with a given angle of incline and radius of curvature, will ensure that the car will remain in its designated path.

[edit] Banked turns with friction

When considering the effects of friction on the system, once again we need to note which way the friction force is pointing. When calculating a maximum velocity for our automobile, friction will point down the incline and towards the center of the circle. Therefore we must add the horizontal component of friction to that of the normal force. The sum of these two forces is our new centripetal force:

{mv^2\over r}= \mu N\cos \theta +N\sin \theta

Once again, there is no motion in the vertical direction, allowing us to set all opposing vertical forces equal to one another. These forces include the vertical component of the normal force pointing upwards and both the car's weight and vertical component of friction pointing downwards:

Ncosθ = μNsinθ + mg

By solving the above equation for mass and substituting this value into our previous equation we get:

{v^2\left(N\cos \theta -\mu N\sin \theta \right)\over rg}= \mu N\cos \theta +N\sin \theta

Solving for v we get:

v= {\sqrt{rg\left(\sin \theta +\mu \cos \theta \right)\over \cos \theta -\mu \sin \theta }}

This equation provides the maximum velocity for the automobile with the given angle of incline, coefficient of static friction and radius of curvature. By a similar analysis of minimum velocity, the following equation is rendered:

v= {\sqrt{rg\left(\sin \theta -\mu \cos \theta \right)\over \cos \theta +\mu \sin \theta }}

The difference in the latter analysis comes when considering the direction of friction for the minimum velocity of the automobile (towards the outside of the circle). Consequently opposite operations are performed when inserting friction into equations for centripetal force and vertical forces.

[edit] Aviation

F-22A Raptor banked to make a left turn
F-22A Raptor banked to make a left turn

When a fixed-wing aircraft is making a turn (changing its direction) the aircraft must roll to a banked position so that its wings are partly angled towards the desired direction of the turn. When the turn has been completed the aircraft must roll back to the wings-level position in order to resume straight flight.[1]

When any moving vehicle is making a turn it is necessary for a centripetal force to act on the vehicle. In the case of an aircraft making a turn, the centripetal force is the horizontal component of the lift acting on the aircraft.

In straight, level flight, the lift acting on the aircraft acts vertically upwards to counteract the weight of the aircraft which acts downwards. During a balanced turn where the angle of bank is θ the lift acts at an angle θ away from the vertical. It is useful to resolve the lift into a vertical component and a horizontal component. If the aircraft is to continue in level flight, the vertical component must continue to equal the weight of the aircraft. The horizontal component is the centripetal force causing the aircraft to turn. (Level flight is flight at a constant altitude.)

Vector diagram showing lift, weight and centripetal force acting on a fixed-wing aircraft during a banked turn.
Vector diagram showing lift, weight and centripetal force acting on a fixed-wing aircraft during a banked turn.

During a banked turn in level flight the lift on the aircraft must support the weight of the aircraft, plus provide the necessary centripetal force. Consequently the lift required in a banked turn is greater than the lift required in straight, level flight. During a turn, the pilot must use the elevator control to increase the angle of attack on the wing to generate the increased lift.

The two forces can be equated mathematically:

{mv^2\over r}= L\sin \theta

where:
{mv^2\over r} is the centripetal force
Lsinθ is the horizontal component of lift
m is the mass of the aircraft
v is the true airspeed of the aircraft
r is the radius of the turn
L is the lift acting on the aircraft
θ is the angle of bank of the aircraft

Douglas DC-3 banked to make a left turn
Douglas DC-3 banked to make a left turn


In straight flight, lift is approximately equal to the aircraft weight. In turning flight the lift exceeds the aircraft weight, and is equal to the weight of the aircraft (mg) divided by the cosine of the angle of bank:

L = {mg\over{\cos \theta}}
where g is the acceleration due to gravity

The radius of the turn can now be calculated: [2]
r = {v^2\over{g \tan \theta}}

This formula shows that the radius of turn is proportional to the square of the aircraft’s true airspeed. With a faster true airspeed the radius of turn is dramatically larger, and with a slower true airspeed the radius is dramatically smaller.

This formula also shows that the radius of turn is inversely proportional to the angle of bank. With a higher angle of bank the radius of turn is smaller, and with a lower angle of bank the radius is greater.

The angle of bank is the sole determinant of the aircraft’s load factor during the turn.

[edit] See also

[edit] References

  1. ^ Federal Aviation Administration (2007). Pilot's Encyclopedia of Aeronautical Knowledge. Oklahoma City OK: Skyhorse Publishing Inc., Figure 3-21. ISBN 1602390347. 
  2. ^ Clancy, L.J., Aerodynamics, Equation 14.9

[edit] External links

[edit] Surface vehicles

  • Serway, Raymond. Physics for Scientists and Engineers. Florida: Saunders College Publishing, 1996.


[edit] Aviation

  • Kermode, A.C. (1972) Mechanics of Flight, Chapter 8, 10th Edition, Longman Group Limited, London ISBN 0-582-23740-8
  • Hurt, H.H. Jr, (1960), Aerodynamics for Naval Aviators, A National Flightshop Reprint, Florida