Is Full Time Use of 4WD a Problem?

[Bluegill2]

I guess I forgot how uncommon common sense is. When I said "unless the situation required / requires it" I figured people would understand things like steep slippery downhill grades, steep slippery uphill grades, slippery side hills, big loader bites that cause too much wheel spin, soft loamy disking that caused too much wheel slip, on and on and on.

Maybe this conundrum is why so many things must now include huge long placards explaining the risks of what used to be common sense knowledge.

In that case 4x4 everywhere on everything, seat belts, arm straps on the stealing wheel, helmets and perhaps we should have to file work plans like we do with airplanes that say where we are starting from, where we are going and when we expect to be finished so somebody will know where to look for us if we are late to dinner.

Ha, that is a good post! It sure does seem common sense isn't as common as it once was...

 

[WaxMan]

Lets make it simple,
Is full time, 4 wheel drive use, a problem Yes or No

NO

 

[TomSeller]

I try to use 2wd, but switch to 4wd when needed..... then forget to switch it back to 2wd.

 

[David_Kb7uns]

People may make any excuse they want for keeping all wheels engaged in 4WD but if it is not needed you will see drivetrain wear increase and so will the wear on the tires.

Many of these smaller CUT and SCUT tractors are not a true 4 wheel drive like the large ag tractors with large front tires and the articulated ones . The smaller tractors sometimes referred to as FWA or Front Wheel Assist where there is an overdrive ratio on the front wheels to keep them from binding and help in the turns. This why they say on hard, dry or with heavy loads to disengage the fronts. Kubota also has the Bi Speed Turn option where the tighter the wheels are turned the faster the front wheels go.

David Kb7uns

 

[ovrszd]

Many of these smaller CUT and SCUT tractors are not a true 4 wheel drive like the large ag tractors with large front tires and the articulated ones . The smaller tractors sometimes referred to as FWA or Front Wheel Assist where there is an overdrive ratio on the front wheels to keep them from binding and help in the turns. This why they say on hard, dry or with heavy loads to disengage the fronts. Kubota also has the Bi Speed Turn option where the tighter the wheels are turned the faster the front wheels go.

David Kb7uns

In the larger AG World, a 4WD tractor is defined as having the same sized tire on all four corners and center pivot. It never comes out of 4WD and never needs to.

Anything other than that is referred to as FWA or MFWD. And needs to be engaged/disengaged as needed/not needed.

 

[ovrszd]

Lets make it simple,
Is full time, 4 wheel drive use, a problem Yes or No

NO

YES.

As in your Avatar.

 

[blueriver]

YES. As in your Avatar.

That's good ..

 

[IgroRockwell]

A few people have expressed concern about overuse of 4 wheel drive due to the smaller size of the front drive train. If you look at the torques involved you can understand that the drive train is adequate. Here is a high-school physics level (simplified) scenario.

Let's say My MX5100 weighs 4000 lbs (no FEL, no loaded tires.) Lets say on a warm, dry summer day a truck leaves the road near me and I hook on to try to pull it out. The tractor is sitting on dry pavement, with dry tires. Lets say that the truck is loaded and, as far as this tractor is concerned, unmovable. But I don't know that yet. I let out the clutch in low and all four tires break free and squawk on the pavement. How much torque is generated by a front tire vs. a rear tire?

Engineers use the term tractive force to represent the force required to break a tire free and start to spin. Tractive force is equal to something called the coefficient of traction multiplied by the weight of the wheel at the ground. On bare asphalt the coefficient of traction simplifies to the coefficient of friction between tire and road. A quick look on the web reveals that a reasonable number for this is 0.75 (no units) or so. If we divide the weight of the tractor by 4 (not accurate, but reasonable) we can assign a weight of 1000 lbs to each tire at the ground. Multiplying, 0.75x1000=750 lbs of force required to break each tire free. (I think it would actually be a bit less on the fronts due to a smaller footprint). Torque on a tire is calculated by multiplying tractive force by the loaded radius of the tire. My fronts have a loaded radius of 1.25 ft and my rears have a loaded radius of 2 ft. Therefore the torque on the fronts is 750lbs x 1.25ft = 937.5 lbs ft. The torque on the rears is 750lbs x 2.0ft = 1500 lbs ft. The torque on the fronts is far less than the rears. In fact, mathematically, torque on unequal sized tires is always a ratio of tire radius. 1.25'(front)/2'(rear)=.625. My fronts can only generate .625 or 62.5% of the rear torque.

In general, without a lot of ballast on the front, as long as the fronts and rears are on the same type of surface, with each tire pulling as hard and efficiently as it can, the fronts can never generate as much torque as the rears. And the smaller the tire, the less torque can be generated. The engineers worked all of this out and designed the front drive system to withstand the maximum torque that the fronts can sustain, plus some overage for when we add weight.

Check it out for yourself. Look into torques on "how stuff works" or some similar site. Or, if you want to really get technical check out this engineering treatise on the subject and see how complicated this all can get.

 

[ovrszd]

A few people have expressed concern about overuse of 4 wheel drive due to the smaller size of the front drive train. If you look at the torques involved you can understand that the drive train is adequate. Here is a high-school physics level (simplified) scenario.

Let's say My MX5100 weighs 4000 lbs (no FEL, no loaded tires.) Lets say on a warm, dry summer day a truck leaves the road near me and I hook on to try to pull it out. The tractor is sitting on dry pavement, with dry tires. Lets say that the truck is loaded and, as far as this tractor is concerned, unmovable. But I don't know that yet. I let out the clutch in low and all four tires break free and squawk on the pavement. How much torque is generated by a front tire vs. a rear tire?

Engineers use the term tractive force to represent the force required to break a tire free and start to spin. Tractive force is equal to something called the coefficient of traction multiplied by the weight of the wheel at the ground. On bare asphalt the coefficient of traction simplifies to the coefficient of friction between tire and road. A quick look on the web reveals that a reasonable number for this is 0.75 (no units) or so. If we divide the weight of the tractor by 4 (not accurate, but reasonable) we can assign a weight of 1000 lbs to each tire at the ground. Multiplying, 0.75x1000=750 lbs of force required to break each tire free. (I think it would actually be a bit less on the fronts due to a smaller footprint). Torque on a tire is calculated by multiplying tractive force by the loaded radius of the tire. My fronts have a loaded radius of 1.25 ft and my rears have a loaded radius of 2 ft. Therefore the torque on the fronts is 750lbs x 1.25ft = 937.5 lbs ft. The torque on the rears is 750lbs x 2.0ft = 1500 lbs ft. The torque on the fronts is far less than the rears. In fact, mathematically, torque on unequal sized tires is always a ratio of tire radius. 1.25'(front)/2'(rear)=.625. My fronts can only generate .625 or 62.5% of the rear torque.

In general, without a lot of ballast on the front, as long as the fronts and rears are on the same type of surface, with each tire pulling as hard and efficiently as it can, the fronts can never generate as much torque as the rears. And the smaller the tire, the less torque can be generated. The engineers worked all of this out and designed the front drive system to withstand the maximum torque that the fronts can sustain, plus some overage for when we add weight.

Check it out for yourself. Look into torques on "how stuff works" or some similar site. Or, if you want to really get technical check out this engineering treatise on the subject and see how complicated this all can get.

Good stuff.

So you always have your tractor in 4wd?

 

[Bluegill2]

A few people have expressed concern about overuse of 4 wheel drive due to the smaller size of the front drive train. If you look at the torques involved you can understand that the drive train is adequate. Here is a high-school physics level (simplified) scenario.

Let's say My MX5100 weighs 4000 lbs (no FEL, no loaded tires.) Lets say on a warm, dry summer day a truck leaves the road near me and I hook on to try to pull it out. The tractor is sitting on dry pavement, with dry tires. Lets say that the truck is loaded and, as far as this tractor is concerned, unmovable. But I don't know that yet. I let out the clutch in low and all four tires break free and squawk on the pavement. How much torque is generated by a front tire vs. a rear tire?

No FEL on the front is the key to your write-up (good stuff BTW). Add the weight of a FEL bucket full or empty and it all changes! I know, have killed a front-end...