"If you decel at 1g with a 4500# car and the 4 straps shift to 15deg to hold it back, the load in each strap is 4500/(sin15* 4 )=4346# per strap, higher if the load is not shared equally (it never is)."
But thanks for checking, I do make lot of mistakes, but I did divide by 4. Adding a pair of longitudinal straps on the rear will add a lot of safety and comfort. But again, I have zero experience transporting a vehicle, which is why I really appreciate the help!
Okay. You guys should quit when you are ahead. We will need to continue to debate until you have valid info for the public from our Forum.
Where is the force of 1g added to the vehicle weight?
Your equation will result in multiplying the sin 15 degrees which is 0.2588 by four which results in dividing just the vehicle weight and not the vehicle weight multiplied by an appropriate factor to represent 1g in your analysis by 1.03. I’ll explain later that mistake is large.
Your fundamental assumption also has all four straps holding the vehicle and tied to the rear of the trailer.
I really don’t understand all the 15 degree shit. 15 degrees off of your force line? That is not crap, don‘t worry about it. Without your precious multiplication by four, the sin of 15 degrees divides the weight of the car by .2588. Did you check your work? Results in nearly 18,000 lbs.
Engineering major and math minor. I still remember a little of it. Sometimes for a living I do have to be the final approval for lifting and moving shit that would make your blood curl. Just trying to help.
Just get a set of pre-engineered straps, and don’t worry about it, you and I don’t design transport systems for a living. And damn I never want to try to achieve -1g with a trailer attached, that would also be quite a feat.