I have a five meter long, 2.40 meter wide, 200 kg portable catamaran I've built at home. To transport, I bought a semi-finished trailer, just the running chassis; and built the upper structure, namely the supports for the catamaran.

Normally they build trailers for a typical mono-hull boats where most of the weight is on the rear side, so they build the chassis such that the wheels and the axle are closer to the end rather than being near the middle of the length.

However, my catamaran is uniform in weight along its length and most of its weight is on the front part (before the axle). Having the axle in the middle seemed more reasonable, so I changed its place:

The Car - Trailer System

I measured the amount of weight exerted on the coupling, it was as low as 30 kgs. Not a problem, but when I tested launching the boat (on the ground), with just about half a meter sliding the catamaran backwards, the trailer's bow took off into the air.

It needed about 10 kgs to come down to the ground. If I stayed on the trailer during the launch it would not be a problem, but this caused me to think of another possibility - possible instability during driving uphill.

Unstable conditions

I tried to formulate the problem in terms of forces and moments. There are two critical pivot points: The trailer wheels (axle) which I believe initiates the unstable action, based on the net moment on the axle. And the coupling which may cause car wheels to take off if the moment is non zero on it.

What maximum incline uphill would cause this instability? Can an equation be provided with a variable axis position so I can optimize the stability using this equation?

  • $\begingroup$ The graph is so unclear that it seems like it could even climb a slope of 90 degrees (if the car is powerful enough), without flipping back. $\endgroup$ Oct 15 '17 at 13:19
  • $\begingroup$ Flipping over : to the left or to the right or backwards. whatever. The maximum slope climeable will be down to the traction available to the tyres... $\endgroup$
    – Solar Mike
    Oct 15 '17 at 13:41
  • $\begingroup$ do you think its important if the car is front or rear drive? $\endgroup$
    – agentp
    Oct 15 '17 at 13:58
  • $\begingroup$ What if Mrear == 10 x Mcar? I dont think even the car will be there with whatever traction power. $\endgroup$ Oct 15 '17 at 14:20
  • $\begingroup$ @agentp that's a godd pıoint which I havent thought of. But I think in case of a flip the rear tires will take off first. Front drive may survive longer :) $\endgroup$ Oct 15 '17 at 14:22

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