New Differential or Rear End for heavy snow in 2wd

[19Walt93]

You never want to stay on top of snow, you want your tires to dig in and push you along. Big contact patch + snow = a slide, and if you're in 4x4 at the time all 4 wheels slide and you're just along for the ride.

 

[scotts90ranger]

Not if you're in 3' of deep snow, then you want to be on top... that's where the big tires aired down to nothing come in. When we go out snow wheeling it's on roads but there's usually a couple feet of snow, go until you can't then air down to ~8psi with 35" tires then go until you can't anymore again then likely turn around... Sometimes there's a foot of snow with a 1/4" of ice on top of that with another foot of snow on top of that, that gets interesting and you do not want to break through that ice layer...

There's many different kinds of snow to drive on, if you're on 6" of snow, sure you want to get to the bottom cuz why not, but if there's ice on the bottom you aren't going anywhere anyway, that's where the aired down siped mud terrain tires come in, they compact the snow around the tread and you can move on ice until you slick it up by too much throttle or too much brake...

 

[rusty ol ranger]

You never want to stay on top of snow, you want your tires to dig in and push you along. Big contact patch + snow = a slide, and if you're in 4x4 at the time all 4 wheels slide and you're just along for the ride.

Actually good snow tires will pack and grab. Digging in the snow is fine as long as you dont either frame out or hit ice down below.

 

[don4331]

Actually the contact patch DOES affect the traction, a great deal. Otherwise racing slicks would be no more effective than bicycle tires.... think about it 'Splain how the size of the contact patch has nothing to do with it......

I grew up in the Ozarks. While we didn't have a lot of snow, we had some of the WORST conditions to drive in. Hills, curvy roads, and a climate that would often cycle between being above freezing during the day, and below freezing at night. Can you say ice? At least one bad ice storm a year.

It's actually easier to drive when it's frozen all the time, trust me.

Anyway - there are theoretical nuances, and then there's the real world. All explained by physics, but there are some factors that are simply of no practical matter. Tread compound, for example, means VASTLY more than where the drive axle is. Learned that once when I bought some GREAT rain tires in the Florida summer.... an inch of the first snowfall of the year back at college, and it was like driving on skis.... Barely made it to the tire store on roads that had the tiniest incline, to get some reasonable tires for the weather.

I've had plenty of FWD and RWD vehicles in snow/ice, and once in a while, mud.

Note:

a. I was merely correcting the formula - Traction = Friction coefficient * weight. There is no contact area in formula.

Part b. of my statement, was Friction coefficient is made up of several variables - contact area being one of them.

But the effect of contact area is a curve- starting a value determined minimum tire size, increasing to an optimum level as you increase contact area, then falling when you increase the optimum. And the optimum varies based on road conditions.​

e.g. Water squishes out from under foot even easier than snow; but while a drag slick will provide optimum contact area on dry, sunny day; it will hydroplane on rainy day. And a grooved tire with lower contact area will work better.​

Snow doesn't "squish" completely out of the way when driving at speed, especially when several inches (feet) deep, wind packed and temperatures well below freezing. And the resistance to the tire packing the snow/lifting the truck up is significant (the 10cm of snow on street in front of house has been packed into ~1cm of near ice and it will slowing increase until March when it thaws). Driven wheels will pull themselves up rather than pushing snow ahead (You can see the piled snow on the right wheel of LoanRanger). In the example @Jim Oaks has presented, given the LoanRanger has a locking rear differential, assuming he could get the truck turned around, back would get him further up the hill.​

​

Temperature is also a variable with a curve. Tire rubber undergoes a glass transition - which is probably why @rusty ol ranger thinks BFG A/Ts are excellent (He is probably driving in temps above 0F); while they go rock hard on me at temps below -5F and I find the Cooper's a better choice. It will also over heat - which is why the Cooper's aren't good in July, but we have 2nd set of tires and rims for my daughter's B4000 and wife's Explorer Sport.​

​

Note: contact area is function of weight in tire and air pressure. Wider tires just make the oval wider but less fore/aft; they don't make it larger.​

 

[rusty ol ranger]

Note:

a. I was merely correcting the formula - Traction = Friction coefficient * weight. There is no contact area in formula.

Part b. of my statement, was Friction coefficient is made up of several variables - contact area being one of them.

But the effect of contact area is a curve- starting a value determined minimum tire size, increasing to an optimum level as you increase contact area, then falling when you increase the optimum. And the optimum varies based on road conditions.​

e.g. Water squishes out from under foot even easier than snow; but while a drag slick will provide optimum contact area on dry, sunny day; it will hydroplane on rainy day. And a grooved tire with lower contact area will work better.​

Snow doesn't "squish" completely out of the way when driving at speed, especially when several inches (feet) deep, wind packed and temperatures well below freezing. And the resistance to the tire packing the snow/lifting the truck up is significant (the 10cm of snow on street in front of house has been packed into ~1cm of near ice and it will slowing increase until March when it thaws). Driven wheels will pull themselves up rather than pushing snow ahead (You can see the piled snow on the right wheel of LoanRanger). In the example @Jim Oaks has presented, given the LoanRanger has a locking rear differential, assuming he could get the truck turned around, back would get him further up the hill.​

​

Temperature is also a variable with a curve. Tire rubber undergoes a glass transition - which is probably why @rusty ol ranger thinks BFG A/Ts are excellent (He is probably driving in temps above 0F); while they go rock hard on me at temps below -5F and I find the Cooper's a better choice. It will also over heat - which is why the Cooper's aren't good in July, but we have 2nd set of tires and rims for my daughter's B4000 and wife's Explorer Sport.​

​

Note: contact area is function of weight in tire and air pressure. Wider tires just make the oval wider but less fore/aft; they don't make it larger.​

Ive driven below zero but never in bad snow on my BFGs...so you may have a point. But i also tow on the freeway heavy shit when its 100* and ive never had one blow out.

 

[MikeG]

Thanks Don, write the formula out for traction - was going from memory and evidently didn't get it right.

 

[alwaysFlOoReD]

I just finished putting on two used winter tires in place of two all seasons on the rear of my 4x4 truck. A definite increase in traction on ice and compact snow. I haven't had them in real snow yet tho. But that's where the 4x4 will come in handy.

 

[MikeG]

I was rolling all this around in my head, and wondered if it was possible to quantify how much force is needed to move a tire through snow.... lo and behold, found this pretty geeky write-up of tests about aircraft tires and how much resistance they have going through snow.

https://www.diva-portal.org/smash/get/diva2:674319/FULLTEXT01.pdf

It's gonna make your head hurt if you aren't up on physics, and especially if you don't do physics in the metric system (which is much easier than inches/pounds/etc.).

While the tire size appears to be similar to various truck tires, they were testing under much faster conditions (minimum 50km/hr or a little over 30mph and up), and under much more load (4,000 Kg or 8,800lbs / tire if I read and converted it correctly), but hey.... they came up with drag figures based on snow depth and density.

So, it ought to be possible to figure out, with a given snow depth/density, and a given slope, just how much force is needed to move the vehicle uphill (with no snow) vs. uphill, with a given amount of snow. The snow drag may well be a much greater percentage than I thought.

 

[MikeG]

I dug through the report a bit more. By chance, one of the reference tires they used (ASTM) tire isn't too far from a standard size of 195/75-r14. There was a second tire tested, and it would appear to be not too different than a 31 or 33" tire that was about a foot wide, but the second tire was tested under much higher load and inflation.

Anyway, the smaller tire was tested at 24psi, and the lowest speed was a little over 30mph. Probably faster than you'd want to be driving in snow, depending on the conditions. Load on the tire was 490 Kg. or 1,078 pounds. If you had 4 tires at that load the vehicle weight would be 4,312 pounds. No too far off if you have 4x4, and maybe a bit of extra weight in the back, although I'd guess most people run larger tires.

So, what is the upshot of this?

Rolling resistance with no snow measured over several runs, averages 125 Newtons per tire, or multiply by 4 tires to get 500 Newtons total.

Add 30 cm of snow (about a foot) of the least dense snow you're ever likely to find (looks like about 3% the density of water), and there's almost no measurable increase in drag, at the lowest speed (~30pmh).
Substitute 30 cm of snow at the max density tested (about 13% that of water) and it appears that the increase in resistance due to the snow adds 42.5 Newtons per tire, or double that if only the two front tires have to push snow out of the way to 85 Newtons. Thus drag increases 17%, which seems significant.... But......

That's just the drag for rolling resistance. It does not count the force needed to go uphill (still working on that). So the difference is going to be even less, percentage wise.

It has been suggested that a driven tire will 'power' its way through the snow, and thus have an easier time moving the snow aside, than a tire that is not driven. I believe that to be exactly correct in mud or sand - because the mud or sand is flung off the back side of the tire, opening up the tread to grab more of it. But I don't think that effect happens much with snow tires, if at all. No doubt we've all experienced tires where the tread filled up with mud, and did not clear out.... those have next to no traction. Snow packs into the tread on a snow tire. Thus, I conclude that the snow tire must compress the snow out of the way, whether driven or not.

If it takes a foot of snow (admittedly not the densest snow you'll ever run through) just to add 17% drag...... then the effect would be marginal, IF the driven tire can compress the snow more easily. But I don't think it can, since the tread stays packed. Add in whatever force it takes to go up the hill, and any theoretical advantage goes right out the window.

There is a difference in drag if you can keep the rear wheels in line with the first. And that difference does go up, the denser/deeper the snow is.

My conclusions.

 

[MikeG]

Force to move the truck up a slope will depend on the angle and mass of the truck.

Mass of truck above (4312 pounds or 1,960kg) times acceleration due to gravity 9.8 m/s = 19,208 Newtons, force exerted on the truck due to gravity. That's what it takes to lift the truck straight up.

Go up a 6 degree slope, as an example, and multiply by the sine of the angle (0.087) so only 1,671 Newtons to move the truck up the slope. Add in rolling resistance (estimated at 500 Newtons above) for a total of 2,171 Newtons. The slope is an inclined plane and makes lifting the truck much easier!

But if the back wheels don't track the front wheels in a foot of fairly light snow.... then we add all of 85 Newtons of drag.... or just under 4%. So if backing up the slope to keep the non-driven wheels in line with the driven wheels makes the difference, then the situation was awfully marginal, one way or another. If you can keep the back wheels in line, it seems that the difference in total force needed would be absolutely minuscule, if it could be measured at all.

Anyway substitute different angles, etc., for your situation. At half the slope the drag makes twice the difference, etc.

Enjoy

 

[19Walt93]

If you've ever walked through foot deep snow you'd notice the drag it causes, wet heavy snow is much worse.
In simpler terms: snow tires with deep tread- good. All season tires or worn tread-bad.

 

[MikeG]

Yes, I have, and yes, you are correct. You can for sure get a LOT of drag from wet, 'slushy' snow, especially when it starts hitting axles, frame crossmembers, suspension parts, etc........

One of the things that jumps out at me with all this - the drag (from the snow) is going to be proportional to vehicle speed (the faster you go the more snow you have to 'squish' in a given timeframe). Drag (force) from snow will be less the slower you go, whereas the force needed to go up a certain incline is independent of the speed (work of course ie. horsepower is not, but force is).

So..... as you slow down, the angle of the incline (even if it is very little) is going to be a greater and greater percentage of the force needed to move the vehicle, vs. the drag from the snow. Thus, at very slow speeds, snow drag gets to be but a tiny percentage of the force needed to move the vehicle.

If that doesn't make sense, then imagine running through a couple feet of snow, vs. walking. The faster you try to go, the more effort it takes, even on level ground.

Since we invariably get 'stuck' as the truck slows to zero, it would stand that the compression force to 'squish' the snow goes to zero, as well.

 

[don4331]

I dug through the report a bit more. By chance, one of the reference tires they used (ASTM) tire isn't too far from a standard size of 195/75-r14. There was a second tire tested, and it would appear to be not too different than a 31 or 33" tire that was about a foot wide, but the second tire was tested under much higher load and inflation.

Anyway, the smaller tire was tested at 24psi, and the lowest speed was a little over 30mph. Probably faster than you'd want to be driving in snow, depending on the conditions. Load on the tire was 490 Kg. or 1,078 pounds. If you had 4 tires at that load the vehicle weight would be 4,312 pounds. No too far off if you have 4x4, and maybe a bit of extra weight in the back, although I'd guess most people run larger tires.

So, what is the upshot of this?

Rolling resistance with no snow measured over several runs, averages 125 Newtons per tire, or multiply by 4 tires to get 500 Newtons total.

Add 30 cm of snow (about a foot) of the least dense snow you're ever likely to find (looks like about 3% the density of water), and there's almost no measurable increase in drag, at the lowest speed (~30pmh).
Substitute 30 cm of snow at the max density tested (about 13% that of water) and it appears that the increase in resistance due to the snow adds 42.5 Newtons per tire, or double that if only the two front tires have to push snow out of the way to 85 Newtons. Thus drag increases 17%, which seems significant.... But......

That's just the drag for rolling resistance. It does not count the force needed to go uphill (still working on that). So the difference is going to be even less, percentage wise.

It has been suggested that a driven tire will 'power' its way through the snow, and thus have an easier time moving the snow aside, than a tire that is not driven. I believe that to be exactly correct in mud or sand - because the mud or sand is flung off the back side of the tire, opening up the tread to grab more of it. But I don't think that effect happens much with snow tires, if at all. No doubt we've all experienced tires where the tread filled up with mud, and did not clear out.... those have next to no traction. Snow packs into the tread on a snow tire. Thus, I conclude that the snow tire must compress the snow out of the way, whether driven or not.

If it takes a foot of snow (admittedly not the densest snow you'll ever run through) just to add 17% drag...... then the effect would be marginal, IF the driven tire can compress the snow more easily. But I don't think it can, since the tread stays packed. Add in whatever force it takes to go up the hill, and any theoretical advantage goes right out the window.

There is a difference in drag if you can keep the rear wheels in line with the first. And that difference does go up, the denser/deeper the snow is.

My conclusions.

I think you missed on your conclusions:

Appendix 1, Table 2 (pg. 45)
Drag for the 7.5-14 under 490kg load, 165kPa pressure, without snow, ~125N agreed

Snow in Calgary is usually about 100kg/m^3, per environment Canada, so I will use closest test values

Add 30mm (a little more than 1") of snow @ 80 kg/m^3 and resistance increases to ~170N - approximately 34% increase.

Increase to 65mm (a little over 2½") of snow @ 80 kg/m^3 and resistance increases to ~245N - almost doubling the force required.

They only do single tests at 70 (2-3/4") and 90mm (3-1/2"), and snow conditions are much heavier (120-130 kg/m^3) but resistance averages 320N. They never do tests deeper (aircraft probably shouldn't be landing in snow deeper; they should wait until runway plowed). I don't feel comfortable extrapolating the data.

Also note in Figure 5.1, that while rolling resistance increases (more/less with speed squared), but there a finite value value for compressing snow at minimal speed - basically, 15cm of 100kg/m^3, will take double force over no snow to initiate motion.

 

[MikeG]

Thanks for the snow density values! Yes denser snow WILL increase the drag due to snow (in a fairly linear fashion) ..... as will the depth of course. But the compression of the snow, due to tires, gets to be a tiny part of the equation, once the snow is a couple inches up the bumper / axles / frame). And the drag on the bumper / axle / frame, won't matter what direction the vehicle is going.

The premise of the question / statement was - basically - will backing up (rear wheel drive vehicle) give substantially less drag in snow, due to the non-driving wheels running in the tracks of the driving wheels? While there are probably many vehicles that can clear a foot of snow, without the axles / frame / etc. adding drag ..... I doubt that there are many that can clear 2 feet of snow. So the amount of snow that needs to be moved (compressed, whatever) goes up a tremendous amount, once the axles/frames contact the snow, no matter what end of the vehicle the drive axle is on.

So in other words, the 2' of snowfall that you reference, isn't going to merely add drag on the tires compressing the snow.... unless you have some sort of really serious lift, and tires well over 4 feet in diameter.

However increasing the density of the snow will add drag, in a fairly predictable fashion, till something besides the tires is contacting snow. But I think most vehicles will maybe hit a foot or so, before the tires aren't the major source of drag due to the snow.

And I still think that a driven tire in snow, won't have substantially different force required to move it through the snow, vs. a driven tire in mud/sand/loose dirt, due to snow not getting flung off the tire.

 

[RonD]

Well a driven tire has an upward vector on any obstacle, even snow, it will pull the tire/vehicle up and over the obstacle, or if its soft enough material, it pulls up/over and will flatten it, no build up infront of the tire
Non-driven wheel can't do that as efficiently so soft material will build up in front of it

And in a RWD backing up a hill can work when going forward didn't, obviously not in every situation, duh
But been there done that quite a few times

Also have experienced Jim's situation with 4x4s open front differential, driven front wheel climbing over the snow, non-driven wheel causing side ways slide as snow builds up in front of it
Makes a good case for cheap lunchbox lockers on the front, lol

It is just physics and can be quantified but like many math proofs they must be changed to meet real world experiments, you can't say "well, the math is correct, so that shouldn't happen", its always the other way around, if it happens the math needs to be corrected to match it
I give you the 3.0l synchro bushings, or the 4.0l SOHC tensioners, "that shouldn't happen......",