Weight Transfer?

[Fallen]

All the talk about corner scales has got me wondering; and maybe someone with corner scales can provide an answer:

When crawling, do the wheel(s) lower down always have more weight on them? Not so much front to rear, as we heavily alter our front to rear weight bias which would affect the answer to my question.

But assuming 50 / 50 left to right balance and all tires in contact with a surface, so you're not hung up on links / bumper / sliders...

During suspension articulation does the higher up wheel have more weight or the lower down wheel? I'm asking to figure out where traction is more likely to be found during suspension articulation; on the higher up or lower down tire?

 

[JatoTheRipper]

Are we assuming the front and rear wheels are contacting the ground at the same angle?

I ask his because the angle at which the tires are contacting the ground would make a huge difference. See my hack of a sketch below. MSPaint FTW.

 

[Gula]

I think this is a very, very tricky physics question... as things change dynamically and drastically in this aspect. How are the scales "laid out" in the scenario? I.E tire placement?

 

[Gula]

Technically the single tire "side" that gives the "torque twist" i, would assume would have the most weight applied to a single scale, in the scenario that Jato and you describe! Right before wanting to flip backwards? I think..... I never took physics.....

However I've worked with a Commercial scale company called intercomp that had a set of 10 like these we used to weigh our 60K "tunner" loaders with, and they had hydraulics for the suspension and just like the toy version it did our vehicle and gave us the center of balance, etc, lol. App controlled, 10 scales at once, 60000 pounds of steel, We always had relatively flat ground, even in the buttholes of the world.....And the vehicle couldn't ever exceed certain dims, as they, we're just it...

Back to reality

Also, if it drove up a per say Z like ramp, similar to what Jato drew, but with 2 "platforms" for the tires to rest, each axle independently hanging, or resting on a "platform" I think the weight from the rig would transfer, but less from the front axle itself. Again this is what I think......

I think the scales would all still require to be parallel in " on a plain" in relation to each other, as in a 3d model per say.... floating in space, rig placed onto scales.....

 

[JatoTheRipper]

I think this is a very, very tricky physics question... as things change dynamically and drastically in this aspect. How are the scales "laid out" in the scenario? I.E tire placement?

Oh definitely a complicated dynamic problem. You can rule out a lot of variables and assume treat it like a static situation.

 

[Gula]

Would be neat to add "rollers" onto the scales, remove the tires and having it on rims only, and then raise the angle of the rig, while maintaining parallelism of the scales comparative to the ground level! Would require building a "jig" OR not. Get 4x those adjustable these: I just found this on AliExpress:
$22.99 | Laser Levels bracket Level Lifting Platform Stand 360 Degree Rotating Laser Support Stand Aluminum Alloy Stand Measurement Tools

https://a.aliexpress.com/_mKtXzid

Imma try it.... buying the scales and these adjustable thingies!

 

[Basicbilly]

As a statics problem, this is a pretty straight forward setup. OP was asking about off camber side hilling with a 50/50 L/R balance.

The tire most directly under the CoG will have the most upward force applied to it.

Assuming:
Static
50/50 L/R balance
All tires in contact with the ground
No articulation of the axles relative to each other
Angle of side hill is consistent
Coefficient of friction same for all tires
Lower tire force has not exceeded friction force
CoG is above ground level

Some of these assumptions are related to each other, but if you start removing them you could find situations where the upper tire could see more force.

You can generally look at the spring compression in any given scenario and let that guide you on what tire will have the most traction. There are tons of situations where my uphill front tire is the only thing pulling me over an obstacle.

 

[Basicbilly]

Also, driving with open differentials will really give you a feel for where the traction is.

 

[Fallen]

Thanks for the interesting replies so far.

I was assuming zero elevation difference front to rear. Driving forward, your driver's side front tire is up on a rock (1 foot tall). Passenger's side front tire is still on the ground.

Which tire has more weight on it? I assume the lower down tire, as it has more weight above it. But maybe the leverage applied through the axle housing changes that?

I assume there's more math involved in this than I can do.

 

[Basicbilly]

At a certain point of complexity, math will not help much, you start looking at it empirically like Gila is thinking about.

Even your last questions has too many unconstrained variables to answer reliably. The lower tire may have more weight on it initially, but if the upper tire has more traction, it will start loading more.

On high traction surfaces, a lot of it is about "hooking" a front wheel to load traction (weight) on it and pull yourself up.

 

[JatoTheRipper]

As a statics problem, this is a pretty straight forward setup. OP was asking about off camber side hilling with a 50/50 L/R balance.

The tire most directly under the CoG will have the most upward force applied to it.

Assuming:
Static
50/50 L/R balance
All tires in contact with the ground
No articulation of the axles relative to each other
Angle of side hill is consistent
Coefficient of friction same for all tires
Lower tire force has not exceeded friction force
CoG is above ground level

Some of these assumptions are related to each other, but if you start removing them you could find situations where the upper tire could see more force.

You can generally look at the spring compression in any given scenario and let that guide you on what tire will have the most traction. There are tons of situations where my uphill front tire is the only thing pulling me over an obstacle.

Writing out the conditions like that...you must be an engineer.

Also, driving with open differentials will really give you a feel for where the traction is.

This is what is super fun about the TRX-4. You can attempt to climb with an open diff and experience some slip. Hit it at a different angle and you'll sometimes find success. Then, of course, there are scenarios where you need the locked diffs.

Thanks for the interesting replies so far.

I was assuming zero elevation difference front to rear. Driving forward, your driver's side front tire is up on a rock (1 foot tall). Passenger's side front tire is still on the ground.

Which tire has more weight on it? I assume the lower down tire, as it has more weight above it. But maybe the leverage applied through the axle housing changes that?

I assume there's more math involved in this than I can do.

Think of it like carrying a couch down a set of stairs. The guy that goes down the steps first will be lower and holding most of the weight while the guy at the top is doing more pulling than lifting.

Good video to explain it in an engineering sense:

 

[Fallen]

Thanks!

It seems @Basicbilly provided a valuable insight for the simple minds like mine.

Whichever shock spring is most compressed is the tire with the most weight on it.

While that's probably where your traction is going to be, there's too many variables (coefficient of friction, torque twist...) for that to be right 100% of the time.

 

[Gula]

Imagine installing a counterweight that travels the edge of the interior of the body, to counter-react going uphills/sidehills!........, or a pendulum that can traverse along a linear rail.... under the body.....to transfer weight to the front, back, sides....lol