Thread: How to calculate Gradeability?

I think you are overthinking it.

A landrover in standard form should have the GEARING to drive any slope fully loaded (note that LR specify a lower GVM for offroad conditions).

What will always be the limiting factor is traction/friction.

To drive a 45 degree slope you need to have perfect friction. So anyone who claims to have driven up a slope 45 degree or greater that is significantly longer than the wheelbase of their vehicle (i.e. too long to use momentum to carry you over) is full of BS.

So in short - IME traction is always the limiting factor in a standard LR, not gearing.

Originally Posted by isuzurover

I think you are overthinking it.

A landrover in standard form should have the GEARING to drive any slope fully loaded (note that LR specify a lower GVM for offroad conditions).

What will always be the limiting factor is traction/friction.

To drive a 45 degree slope you need to have perfect friction. So anyone who claims to have driven up a slope 45 degree or greater that is significantly longer than the wheelbase of their vehicle (i.e. you can't use momentul to carry you over) is full of BS.

So in short - IME traction is always the limiting factor in a standard LR, not gearing.

Yes Ben I know that traction will be the limiting factor, unless overloaded, a very tired engine and/or too large a tyre are part of the equation,and i've run out of power on a few occasions for all the above. but that doesn't really answer my question.There are mathematical formulas on the net for calculating climbing abilty but the way they are laid out makes them difficult to follow for this old codger. I'll give you an idea why I would like to know. Unimogs are criticised for having too much antisquat geometry in their torque tube suspension design. If I want to build a vehicle with Unimog axles and a 4ft long one link wishbone to replace the torque tube, I can calculate how much lifting force is applied via the wishbone pivot to the chassis at mid wheelbase for a given amount of axle torque. if I have a formula for calculation how much axle torque is required to crawl up a more realistic gradient of say 30 degrees, lets say 4000lbft, then with a 4ft long wishbone I know that the lifting force would be 1000lb, which may in the antisquat scheme of things on avehicle weighing 4000lbs be equivelant to two tenths of bugger all.
Yes I'm ovethinking things, and I don't really NEED to. But it's something that has interested me of late, and a little mental excercise might keep this old brain functioning a bit longer.
Wagoo.

I've always used the formula as follows:

I can't get up = It's too steep or too slippery

Who in the world would try to work it out using a mathematical formula?

Originally Posted by wardy1

I've always used the formula as follows:

I can't get up = It's too steep or too slippery

Who in the world would try to work it out using a mathematical formula?

Good one.
Filed for future reference.

Originally Posted by wagoo

Yes Ben I know that traction will be the limiting factor, unless overloaded, a very tired engine and/or too large a tyre are part of the equation,and i've run out of power on a few occasions for all the above. but that doesn't really answer my question.There are mathematical formulas on the net for calculating climbing abilty but the way they are laid out makes them difficult to follow for this old codger. I'll give you an idea why I would like to know. Unimogs are criticised for having too much antisquat geometry in their torque tube suspension design. If I want to build a vehicle with Unimog axles and a 4ft long one link wishbone to replace the torque tube, I can calculate how much lifting force is applied via the wishbone pivot to the chassis at mid wheelbase for a given amount of axle torque. if I have a formula for calculation how much axle torque is required to crawl up a more realistic gradient of say 30 degrees, lets say 4000lbft, then with a 4ft long wishbone I know that the lifting force would be 1000lb, which may in the antisquat scheme of things on avehicle weighing 4000lbs be equivelant to two tenths of bugger all.
Yes I'm ovethinking things, and I don't really NEED to. But it's something that has interested me of late, and a little mental excercise might keep this old brain functioning a bit longer.
Wagoo.

OK - so you want to calculate anti-squat - a different question. I am no expert on that, but plenty on ol and pirate are. have a read here:
Outer Limits 4x4 Board • View topic - Anti Squat, anti Dive suspension geometry.
(and associated links). Be forewarned - lots of great tech but you need a thick skin on pirate.

From my limited understanding, A single-link suspension like you describe, terminating at the CofG of the vehicle will give 100% AS.

Originally Posted by wardy1

I've always used the formula as follows:

I can't get up = It's too steep or too slippery

Who in the world would try to work it out using a mathematical formula?

If you've got nothing useful to contribute then kindly move along. There's nothing of interest here for you.
Wagoo.

Originally Posted by isuzurover

OK - so you want to calculate anti-squat - a different question. I am no expert on that, but plenty on ol and pirate are. have a read here:
Outer Limits 4x4 Board • View topic - Anti Squat, anti Dive suspension geometry.
(and associated links). Be forewarned - lots of great tech but you need a thick skin on pirate.

From my limited understanding, A single-link suspension like you describe, terminating at the CofG of the vehicle will give 100% AS.

Thanks for the link Ben.That bloke daddylonglegs seems like he's up himself
Wagoo.

Originally Posted by wagoo

Thanks for the link Ben.That bloke daddylonglegs seems like he's up himself
Wagoo.

Bill/daddylonglegs/Agrover/portalrover knows his stuff and has built some of the most innovative landies around - home made portals, home made underdrives, forced articulation, a 6x6 IIA with great wheel travel, etc etc.