From another long trailer hauler who has the math down better than I do. This trailer is different than yours but answer should be in the ball park.
Assumed 40 ft long, 10 ft wide ag trailer with pivoting front axle and rigid rear axle being pulled by pickup truck:
“OK here's what I've come up with. I'll assume that the road makes a 90 degree arc around a point I'll call X. I'll assume the rear axle must be perpendicular to the trailer and that the front axle pivots. When the trailer is turning, the front and rear axles will fall on radi of the circle it is turning on, in our case that circle's center is at X. For the purposes of making a clear mental picture I will assume a right turn (the outcome will be the same for either right or left). The part of the trailer closest to X will be the right rear wheel, I shall call the distance from the right rear wheel to X, A. The furthest point on the trailer from X will be the left front corner, I shall call the distance from the LF corner to X, B. The width of the road is 20' so for the tightest possible turn... B-A=20' Add A to both sides B=A+20' The width of the trailer is 10' and the length is 40' so in order to get from the RR wheel to the LF corner I have to go outward 10' and forward 40'. Now I construct a right triangle with one leg being A+10', the second leg being 40', and the hypotinuse being B. (A+10)2+402=B2 Substitute A+20 for B ((A+10)2+1600)=(A+20)2 multiply out the squares A2+20A+1700=A2+40A+400 Subtract A2+20A+400 from both sides 1300=20A Devide both sides by 20 65=A So if the distance from the center of the turn to the inside edge of the road is 65' or more then the trailer can make the turn.”
