Relation Between Turning Radius and Speed of Vehicle

Two factors:

• Wheel grip with the surface over which it is passing
• Comfort of the occupant

http://en.wikipedia.org/wiki/Handbrake_turn

http://en.wikipedia.org/wiki/Cant_(road/rail)

http://en.wikipedia.org/wiki/Wall_of_death_(motorcycle_act)

Centripetal acceleration a = v²/r; therefore, r is proportional to v².

Depends on the car, grip (tires), camber, roadsurface type and condition, or is this purely a theretical question?

but can u give me actual method or any type of formulae to calculate turning radius (while turning)of vehicle at a particular speed........?

This is simple enough for an 8 year old child to do. Perhaps you should fetch an 8 year old child?

Being childish, myself, I guess I will have to do.

a = v^2/r

You will need to know the maximum lateral acceleration of the car.

Solve for v:

Rearrange the equation as follows:

v = √(a * r) <== Basic algebra

Make sure your units match (i.e., if you use meters per second, your radius must also be given in meters).

the turning radius of vehicle at different speed is depend on the steering angle slip angle also related to mass of vehicle,wheel base and wheel track...

Thus i hav to calculat turning radius on those parameter...!!!!!!

As long as you do not exceed the corner grip of the tires (i.e., begin to slide) the calculation is simply the average of all four wheels.

If you want to complicate it a bit more, then you need to consider the roll angle of the vehicle (side to side) and the contact patch of each tire on the surface, since the outside wheels bear more downforce than the inside wheels on the road surface while turning.

However, you originally framed the question as a simple "relation between turning radius and speed of the vehicle". The work I did in the previous post is correct for that question and the vehicle itself. It assumes the car is not sliding, but performing at the edge of its lateral G limit or at some specified G load below that limit.

If the vehicle is in a slide, the coefficient of dynamic friction is less than the coefficient of static friction and your turn radius would suffer greatly. That is, you would not be able to maintain the same velocity to maintain the same turning radius (velocity would be less and so would your G load). That is why in racing it is faster to take turns without going into a slide or "drifting" as the term is called unless it is a Hollywood movie.

Dear Mr. Epke,

Pl. see my posting No. 16 below.

DHAYANANDHAN.S

The equilibrium cant height (the outer rail higher than the inner) for a 4ft 8¹/₂in gauge rail vehicle is 0.0562v²/r, where r is the radius in chains, v is the maximum line speed in mph and the cant is in inches. It is usual to put on 2/3 of the equilibrium cant to reduce rail and wheel wear around the curve. There are tables of cant deficiency to limit discomfort around curves at speed, and these can be used to determine the maximum speed over a connection into or out of a canted curved line.

The maximum cant on any British railway line is reckoned at 6in.

For futher reading try the chapter on curved track in "British Railway Track, design, construction and maintenance", any edition, published by the Permanent Way Institution. Metric equivalent measurements apply to the later editions.

Why are folk here talking about railways?

the turning circle is determined by the steering components. Increasing or decreasing the speed cannot alter the turning circle.

As to whether the vehicle remains upright, continues in a straight line or simply skids a bit is down to the relationship between the forces downwards / tyres friction etc and centrifugal forces trying to push the vehicle outwards.

Let the passenger get the train - that their G force issue resolved.

Increasing or decreasing the speed cannot alter the turning circle.

Sure it does. That's why he's asking, I would assume.

Given a fixed steering wheel angle and a particular speed, a certain side force will be generated. Based upon this side force, the relative tire pressures, etc, etc, etc, etc, etc, etc, the front tires and rear tires will operate at different slip angles. (This is the case with most road vehicles vehicles most of the time.) For most road vehicles, as you accelerate around a skid pad, you must feed in more steering angle to maintain a constant radius.

Changing speed changes the side force, which changes the slip angles and the roll attitude of the vehicle, and therefore wheel/road geometries and both side-to-side (dramatically) and fore-and-aft (slightly) weight transfer.

Other things being equal (but they are not) if the slip angle at the front increases more than the slip angle at the rear, the car will tend toward a larger radius. The change in roll from the change in side force cause changes in camber in different ways at the front and rear. Bump steer changes the relationship between steering wheel angle and road wheel angle. The rear end geometry may induce roll steer.

More simply, all the factors that cause one car to oversteer and another to understeer also affect the degree to which the turning circle changes with speed given a constant steering wheel position.

Calculating the change is not possible by the use of any simple formula with a just a couple variables. Such changes are instead modeled, given all the relevant vehicle and tire traits.

The geometry and force interactions were discussed here previously.

Correct - they were. My point is that there is no relationship between speed (and thats not the correct term) and turning circle.

That's right.

There is a relationship between the slip angle and the peripheral velocity, where the slip angle is dependent on the tyre friction and vehicle mass.

OP needs to draw some vectors to come up with a unifying formula.

Sounds like work to me.

Dear Mr.yadaosurajd

This subject/topic is covered in the 2nd year Mathematics during our Study days in Engineering Collge. This is covered in the Mathematics - DYNAMICS chapter, under the Caption - "BANKING of VEHICLES".

The Equation governing the SPEED of the Vehicle and the RADIUS of the Road is as follows:

tan (Theta) = V^2/(R x g) where V is the Speed of the Vehicle, in Metres/Sec., R is the Turning Radius of the road while turning, g is the Acceleration due to gravity which is 9.81 Metres/Sec^2. ( OR)

Inclination of the Road to Horizontal in Degrees = tan (inverse) V^2/ (R x g)

This equation will indicate that the Speed of the Vehicle and Radius is linked by this Equation and the inclination of the road will vary as rhe speed varies or Raduis varies, and for a given speed and radius the inclination is constant.

DHAYANANDHAN.S

Note: The Symbols in my computer is not working, hence tan-1 is shown as tan (Inverse)