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Effective weight of trailer with wind drag calculated in.

bucolic

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So I was just wondering if there are any engineers out there that might know how wind drag factors into the effective trailer weight. We all know that towing say a flatbed trailer with a car on it is much different than towing a camper that is 10' high with a large surface area. I know there are also endless variables that come into play on this such as headwind speed, side wind, vehicle speed, etc but was just curious if this has been calculated out before?

So for simplicity say you have are towing a camper that weighs 6000 lbs fully loaded. It is 10' high 8, wide, 26' feet long. Towing on a flat surface at 70mph would be like towing an XXXXX lb trailer with no drag. It seems to me it would be like the tractor pulls with the sliding weight. The faster you go the higher the drag and the effective weight of the trailer would increase with speed. I know when I first pulled my camper I took off out of the driveway and thought wow! This thing is a breeze to pull I don't even know it's back there. Once I hit 50MPH with a 20MPH headwind I couldn't get out of my own way!

I apologize if this has already been answered before but I didn't see anything when searching.
 

Sascwatch

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You would have to know the square footage of the front of the trailer to calculate the aerodynamic drag, added weight due to a sloped front is way above my physics knowledge but I’m sure it’s possible to calculate it knowing the speed, size and angle of the front.
 
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brian42

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I don't know if you would be able to generalize the equation as there are too many individual variables for each situation (frontal area, aerodynamic shape, material, is it waxed, etc.).

Looks like someone else had this question too a few years ago:

 

LaxDfns15

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You can get a *very, very* generalized number, but you're never going to get a real-world value that will help you.

Fd = Drag Force
C = Drag coefficient. Typically changes with velocity, but assumed constant a lot of the time. This is based on the shape of the item.
p = density of fluid you're moving through
A = area of object moving
v = velocity of object

Fd = 1/2*C*p*A*v^2

So taking some numbers from your example trailer with no wind break in front of it (truck in front of the trailer changes things immensely), we've got this

C = 1.17 for square plate perpendicular to movement
p = 0.0765 lb/ft^3 density of dry air at sea level (density changes from elevation, temperature, humidity, etc. tons of factors)
A = 8 ft * 10 ft = 80 ft^2
v = 70 mph = 102.67 fps

Fd = 1/2*1.17*0.0765*80*102.67^2 = 37,739 pounds of drag at sea level, in dry air, with no truck in front of it.
 

bucolic

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I don't know if you would be able to generalize the equation as there are too many individual variables for each situation (frontal area, aerodynamic shape, material, is it waxed, etc.).

Looks like someone else had this question too a few years ago:

Good reading in there! I guess Airstream trailers may be easier to pull just based on the rounded corners!
 

Dragonmaster13

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Either way it doesn’t matter as most truck can pull WAY more than they can stop, think tundra pulling the space ship, f-150 pulling a train... they can get them going but stopping is another story.


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