F250 and up, max 60 square feet frontal area

TucsonJim wrote:

Ron Gratz wrote:

TucsonJim wrote:
A simple trigonometry equation would be if you had exactly 60 square feet and it was sloped at 15 degrees. 90 degrees (perpendicular to the air flow)- 15 degrees slope = 75 degrees. Sin of 75 degrees is .966. Multiply 60 square feet by the inverse of .966 to get the equivalent square foot load. In other words, the recommended limit is 60 square feet at 90 degree impact angle, but 62 square feet at a 15 degree slope.

That depends on how they define "frontal area".

Usually, "frontal area" means the projected cross sectional area which, for your example, would be 60 sq ft for both the vertical front and the sloping front.

Ron

You're right Ron. That's why I used a simple evaluation if everything was at a 15degree slope. When you put all the rounded corners, slopes, etc. of the TV and the TT/FW, it gets quite complex.

Like Ron said, "Frontal area" ignores the slope. If you drove the rig through a cartoon brick wall, what is the area of the hole left in the wall? That is frontal area. Aerodynamic drag is the frontal area multiplied by the coefficient of drag (Cd). Round corners and sloped front will decrease the Cd and so effect the drag - but not by changing the frontal area.

It is interesting that the manual lists "frontal area" and then comments on factors effecting Cd. The load on the tow vehicle is caused by the total drag, not the frontal area. Oh, and the other factor in calculating total drag is airspeed. Frontal area times Cd times airspeed squared.

Personally, I would not pay much attention to what the manual states regarding frontal area because it is incomplete information since it doesn't mention Cd or airspeed.