Homework Help: Understanding drawbar pull coefficient: v basic

1. Apr 9, 2013

hooleydooley

1. The problem statement, all variables and given/known data

Hi, this is not homework or course work, I am a layman and I am just trying to understand an article I have regarding drawbar pull, in particular the pull coefficient (dimensionless) P/W where P is described as "pull (drawbar pull), lb" and W is described as "wheel load; weight, lb".

2. Relevant equations

The equation seems simple enough but the problem arises when the P (drawbar pull) seems to be measured in horizontal force and the W (wheel load; weight, lb) seems to be measured in weight, then it makes no sense to me.

3. The attempt at a solution

There are 3 solutions I can think of:

1) The carriage attached to the wheel in the test is measuring weight, although at one point n the paper it says "horizontal force"

2) The "lb" in "W wheel load; weight, lb" refers to the horizontal force required to propel the wheel at a given speed/acceleration.

3) A pull coefficent of .5 does not mean that the wheel can only tow 50% more weight*, or with 50% more force than it takes to dive the wheel, before traction is lost.

*assuming the same rolling resistance as the wheel.

I have searched the net extensively but because most drawbar pull formulas are related to locomotve power they don't help to clear things up for me, and I need to be sure that the pull coeffcient is representative of a pulling power or traction in relation to the wheel (ie: pull coefficient of 0.5 equals 50% towng capacity above either the wheel weight, or the horizontal force require to drive the wheel).

There's a big difference between a wheel with 1000kgs of weight being able to pull 500kgs of weight (assuming the same rolling resistance as the wheel), or being able to pull with a horizontal force of 500kgs, before traction is lost.

I realise this question is very lowbrow for this forum but I am hoping a definitive answer can be given to help me out.