I have a question for anyone who owns Carroll Smith's excellent book.
Please open your text and turn to page...er...section 2-15
The diagram of the traction circle
Now it illustrates a tire that has a maximum lateral and longitudinal limit of 1.5 gs. Now based on where the line ends the tire is operating at 1.1g cornering force and .8g acceleration force. This is also backed up by a notification in the greyed text. Now, based on my primitive math that adds up to 1.9g.
From the book "In other words, the tire can simultaneously generate an amount of braking thrust and an amount of cornering force which, added together, will total more force than the tire is capable of generating in any one direction"
I dont get this. The entire point of the traction circle is that if you are using 1.1g of a 1.5 G tire for cornering, you only have .4g left for acceleration. How did we manage .8g?
PLEASE explain in the simplest plainest english possible. I'm very much a visual learner and need to be able to form this concept in my head.
Also, when he mentions "rear droop travel" is he referring to rear rebound in the dampers?
Ross Stonefeld
Aztec Group Motorsport[p][Edited by Ross Stonefeld on 08-08-2000]
Drive to Win
Started by
Ross Stonefeld
, Aug 08 2000 03:28
2 replies to this topic
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#2
Ursus
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- 2,411 posts
- Joined: March 99
Member
Posted 08 August 2000 - 09:44
Ross, I don't have the book but I think I can help you anyway.
When you are adding forcesyou need to consider their direction aswell as their size. E.g. two equal size forces but with opposite directions (and acting on the same point) will cancel each other out.
In your case you have two forces, one acting forward and the other sideways. When you add these you will get a resulting force with a direction somewhere inbetween the others and a size not quite the sum of the others.
An easy way to graphicly get both size and direction is to draw the forces as the sides of a rectangle, then the diagonal is both the right size and has the same direction as the resulting force. (See pic.)
Hope this helps.
When you are adding forcesyou need to consider their direction aswell as their size. E.g. two equal size forces but with opposite directions (and acting on the same point) will cancel each other out.
In your case you have two forces, one acting forward and the other sideways. When you add these you will get a resulting force with a direction somewhere inbetween the others and a size not quite the sum of the others.
An easy way to graphicly get both size and direction is to draw the forces as the sides of a rectangle, then the diagonal is both the right size and has the same direction as the resulting force. (See pic.)
Hope this helps.
#3
tak
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- 354 posts
- Joined: November 98
Member
Posted 11 August 2000 - 00:52
Ross-
Ursus is correct in her response. The mathmatics is the pythagorian theorem: if we call total traction "T", lateral acceleration "X" and longitudenal acceleration "Y", then
T^2 = X^2 + Y^2.
The rectangle in Ursus's diagram has 2 right triangles. In each triangle, 1 leg is lateral acceleration, the other leg is longitudenal acceleration. The Hypetenuse is the resultant force.
Ursus is correct in her response. The mathmatics is the pythagorian theorem: if we call total traction "T", lateral acceleration "X" and longitudenal acceleration "Y", then
T^2 = X^2 + Y^2.
The rectangle in Ursus's diagram has 2 right triangles. In each triangle, 1 leg is lateral acceleration, the other leg is longitudenal acceleration. The Hypetenuse is the resultant force.
