# Rotational frictions relationship to surface area

1. Jul 8, 2014

### pirsquared

Hi,

I have a question I'd like some help with which is related to my job working with drill rigs.

Where you have a cylinder such as a drill string where the forces of the ground pressing in from all sides how much influence does the surface area have on the rotational force required to keep turning the cylinder e.g. does a cylinder 20m long and 150mm wide need 20% more rotational force to turn than a 125mm cylinder if the pressure from the surrounding ground is equal on both

thanks

2. Jul 8, 2014

### 256bits

Well, the radius ( diameter ) is 20% greater going from a 125 mm to 150 mm wide cylinder, so what is your justification in saying that the torque ( rotational force ) would then be 20% greater? ( Hint: the circumference is also greater. )

3. Jul 9, 2014

### pirsquared

Well there's more surface area in contact with the surrounding ground so I was wondering if the additional force required to rotate the pipe is proportional?

4. Jul 9, 2014

### A.T.

I think the frictional torque it will be proportional to the square of the radius. Assuming the same pressure, friction is proportional to the area. The torque is proportional to radius and to the force, which here itself is proportional to radius.

Of course, if you drive the drill by applying forces at the cylinder circumference, the forces needed will be simply proportional to the radius.

Last edited: Jul 9, 2014
5. Jul 9, 2014

### pirsquared

Is the above assuming that you are always turning the tube from one point on the outside of the cylinder similar as to if you were turning a steering wheel as in this case the larger diameter is an advantage. What happens in the case where the driving force is at the center of the cylinder (lets assume a pipe 100mm in diameter) and rotating firstly the 125mm pipe and then the 150mm pipe under the same frictional forces. Does that affect the outcome

6. Jul 9, 2014

### A.T.

As I said: If you drive the drill by applying forces at the cylinder circumference, the forces needed will be simply proportional to the radius.

As I said: The frictional torque it will be proportional to the square of the radius. So if the driving forces are always applied a the same radius, they will also be proportional to the square of the radius.

Last edited: Jul 9, 2014
7. Jul 9, 2014

### 256bits

As above for dynamic friction.
Except what is the frictional force due to? This we do not exactly know.

If it is regular dynamic friction, or sliding friction of one part against another, that depends only on the coefficient of friction of your pipe and the material your are drilling into. Here, the torque you have to supply to turn the drill is proportional to the radius squared, due to a larger surface area under pressure producing more fiction force, and this friction force acting on a larger area.

If it is viscous friction, which it could be since a drill rig would use a type of mud as a lubricant, then the torque to be supplied could be a function of radius cubed. The reason being is that viscous friction is proportional to the velocity between the relative velocity of the moving parts on either side of the fluid. With a larger drill radius, the circumference of the drill is moving more quickly. You thus have a larger area subject to friction and the larger area moving at a greater velocity.

You will have to determine the type of friction the drill is subject to.
These discourse if for ideal conditions by the way. Out in the field there could some other friction variables at play, but you do have the idea now, OK.

PS,
Adding the Wiki for viscous friction
http://en.wikipedia.org/wiki/Viscous_friction