Understanding drawbar pull coefficient: v basic

[hooleydooley]

Homework Statement

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".

I should point out that this article is regarding traction and not power.

Homework 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.

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

 

[KataKoniK]

Hello,

Thank you for your question. I understand your confusion regarding the units of P (drawbar pull) and W (wheel load; weight). In order to understand the pull coefficient, it is important to have a clear understanding of the units being used.

First, let's define the terms "pull" and "weight". Pull refers to the force required to move an object, while weight refers to the force of gravity acting on an object. Both of these are measured in units of force, such as pounds (lb).

Now, let's look at the units for the pull coefficient (P/W). As you correctly pointed out, P is measured in units of force (lb), while W is also measured in units of force (lb). This means that the pull coefficient is a dimensionless quantity, as the units of force cancel out when divided.

In simpler terms, the pull coefficient is a ratio of the force required to move an object (P) to the force of gravity acting on that object (W). This means that a pull coefficient of 0.5 does not mean that the wheel can only tow 50% more weight, but rather that it requires 50% of the force of gravity acting on the wheel to move it.

In terms of your three proposed solutions, I would say that the first and third solutions are not correct. The carriage attached to the wheel is not measuring weight, but rather the force required to move the wheel. And the pull coefficient is not representative of a pulling power, but rather a ratio of forces.

I hope this helps to clear things up for you. Please let me know if you have any further questions or if I can provide any additional clarification. Best of luck in your understanding of drawbar pull and traction.