Wheel ground pressure vs wheel size

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Summary
Ground pressure from a wheel - knowing the max load on a small wheel, how big must a wheel be to carry a large load?
I need to drive a tractor across and area of soft ground without leaving deep tracks. I tested the ground carrying capacity with car and concluded that 700kg is max load on a 265/70-17 wheel, with 2,2bar tire pressure. (That wheel is 265mm wide, 401mm outer radius, unloaded.)

As the ground is soft, the tire threads will sink in. Hence I don't think we need to account for thread pattern.

The tractor (with load) weights 3500kg on the most loaded wheel. I need to figure out how big tyre I should buy for the tractor tire in order to get similar ground pressure as the car tire. Tire pressure should not go below 1,1bar because of risk of tire falling off the rim in terrain.

Available tire widths are 520mm, 600mm, 650mm and 700mm. For these widths, which radius is required?
(Tire cost increases exponentially with size)

The main challenge in this question is calculation of the tire contact area from the wheel dimensions and pressure. When a 265mm wide tire carries 700kg, a 600mm tire of the same radius and pressure will carry (700kg * 600mm/265mm = ) 1585kg. I need a contact area to be 2,2 times this in order to carry 3500kg. How much will the contact area increase when tire pressure is down from 2,2bar to 1,1bar? How much will the wheel radius need to increase to the rest of the needed area?
 

Baluncore

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The ground pressure of a tyre is simply the tyre pressure.

If you know the axle weight of the car and the car tyre pressure, you can work out the ground contact patch area for each wheel.

The tractor has a lower tyre pressure than a car, so it will always have lower ground pressure than the car. The tractor tyre size and pressure was specified by the manufacturer to support the weight of the tractor without the rim damaging the tyre sidewall against the ground.

The wheel slip when drawing a load will be the limiting factor. What is the ground slope?

Why do you need to cross a bog? Are you pulling a load or an implement?
 
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If ground pressure = air pressure, how can it be that big machines (that usually uses low tire pressure) makes deeper tracks in the ground than smaller machines? Are there other factors I need to pay attention to?
 
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If ground pressure = air pressure, how can it be that big machines (that usually uses low tire pressure) makes deeper tracks in the ground than smaller machines? Are there other factors I need to pay attention to?
Low pressure yes, but also bigger area in contact with the ground.

pressure (pounds-per-square-inch) times area (square-inches) gives lifting force (pounds).

Think of a bulldozer with a tank-tread. It has much more tread area touching the ground than the bottom of a tire.
 

JBA

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For these widths, which radius is required?
What Larger Standard Tire Diameters are available?
 
147
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some alternatives:
600/70-30: 600mm wide, 801mm radius
650/60-34: 650mm wide, 822mm radius
710/60-34: 710mm wide, 858mm radius
540/60-30: 540mm wide, 732mm radius
480/60-28: 480mm wide, 668mm radius
520/60-34: 520mm wide, 796mm radius
520/60-38: 520mm wide, 846mm radius
620/60-42: 620mm wide, 967mm radius (this may not have enough space)
 
147
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Low pressure yes, but also bigger area in contact with the ground.

pressure (pounds-per-square-inch) times area (square-inches) gives lifting force (pounds).

Think of a bulldozer with a tank-tread. It has much more tread area touching the ground than the bottom of a tire.
Lowering pressure increases contact area, so it does make sense. But large radius sinks in less than small radius. How does that factor come into this picture?
 

Baluncore

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But large radius sinks in less than small radius. How does that factor come into this picture?
The contact patch has an elliptical outline. Large diameter wheels have a longer ellipse, wider tyres have a wider ellipse. The area of an ellipse is proportional to the product of length and width. A = Pi * a * b.

If ground pressure = air pressure, how can it be that big machines (that usually uses low tire pressure) makes deeper tracks in the ground than smaller machines? Are there other factors I need to pay attention to?
Cars do not drive on soft ground because they sink in, then the drive wheels dig holes by shearing and displacing soil. Cars do not leave deep tracks on soft ground because they cannot get there.

Tractors are supposed to “float” over the top of the soil, but require a significantly larger contact patch to support their greater weight, especially with the lower tyre pressure.

Traction requires that horizontal plane soil shear not be exceeded, which requires a large contact patch area and explains why tracked vehicles have an advantage pulling on soft ground. For traction on very soft ground a tractor may be fitted with wider tyres or with dual rear tyres, but the tyre pressures must be kept low.

Until we know your situation better we cannot advise you on what other factors you need to consider.
 
147
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So the only place the factor of larger radius wheel comes into the picture is that small radius wheels needs to deform more in order to create the same contact area?
Maybe that deformation will cause the tire to "climb" more when rolling and that climb (process of deforming the part of the tire that is on the way down) rips the bindings in the ground a bit apart, then the ground is weaker when the weight comes?
 

JBA

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After reviewing the factors affecting your issue of tire size vs load vs pressure, the bottom line seems to be that the standard formula used for determining a tires area of contact on a hard flat surface equals load/tire inflation pressure; however, this does not indicate what size of tire is capable of carrying that load or providing that calculated contact area for a given inflation pressure. For that issue you are best served by consulting with your tractor tire supplier.
 

Baluncore

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Maybe that deformation will cause the tire to "climb" more when rolling and that climb (process of deforming the part of the tire that is on the way down) rips the bindings in the ground a bit apart, then the ground is weaker when the weight comes?
Your term “climb” is confusing. As a tyre rolls on soft ground it compresses the ground to make a depression. That increases the density of the ground until it can bear the tyre pressure, or the vehicle will sink in to the axles.
It is easy to think of the ground depression as being a hole that the wheel must climb out of, but really both ground and tyre are deforming without change in height. It takes more energy to compress the ground once than to reverse and retrace the track, which implies that moving forward over new ground generates more ground shear.

You refer to a tractor but do not define the load being drawn or the traction force required that must be carried by ground shear. Efficient traction requires there be some wheel slip, something like 5%, but that assumes one passage say during cultivation. If you have less than 5% wheel slip you need a bigger tractor, a smaller plough, or you should be driving faster to waste less time.

Tyre type can be a belt like a road tyre, or a balloon like a tractor. The tread can be designed to displace water from the road contact, or to clean surface mud from the ground contact patch like the tread on an agricultural tractor. A car tyre has a more rectangular contact patch than the ellipse of a balloon tyre.

The interpretation of a tyre specification such as 600/70-30 tells it all.
The tyre tread is 600 mm wide, and it fits to a 30 inch rim.
The profile is 70%, which means the height of the rubber is 70% that of the tread width.
The outer diameter will be; D = (2*600*0.70)+(30*25.4) = 1602 mm, radius = 801 mm.
The rolling circumference will be C = Pi*D = 5033 mm, which may have gear selection implications.

The contact patch will be 600 mm wide. Half width is; b = 600 / 2 = 300 mm.
Given a crude assumption that a tractor tyre will be deformed to half it's profile height before the sidewall contacts the ground, you can calculate the maximum patch area from simple geometry.
The unloaded rim sits 600*0.70 = 420 mm from the outer tread. When loaded that falls to half, d = 210 mm. The half length, of the contact patch is half the length of a chord, 210 mm deep, on an 801 mm radius circle; a = √(r2 – (r-d)2) = 540.6 mm.
The patch area will be the area of an ellipse; Pi*a*b = 0.51 m2.
Given a tyre pressure of about 12 psi = 80 kPa the force is 80k * 0.51 = 40,800 newton.
Given g = 9.8; Supported mass is; (40800 / 9.8) = 4,163 kg maximum per wheel.

For a particular tread width, a higher profile gives more clearance between the rim and the ground, which allows a lower pressure, or reduced sidewall damage on rough ground.

Wider rims are needed with wider tyres to control excessive sideways vehicle movement.
When running dual wheels the sidewalls of the two tyres should not contact at the bulge near the ground.

You need to identify the ground conditions, muskeg bog, desert dune, or gravel beach, and define the traction requirement and ground slope before you get a reliable understanding of the situation. Talking generalities does not really help a particular case where identifying specifics is critical to understanding that case.
 
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cjl

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The ground pressure of a tyre is simply the tyre pressure.
Unfortunately, it isn't that simple. Tires themselves have a structural strength, so ground pressure is typically a bit above (or sometimes a lot above) tire pressure on a hard surface. In addition, on soft ground, if the ground deforms around the bottom of the tire, the ground pressure can be substantially below the tire pressure. You can't really know what the actual ground pressure is without a pretty detailed model or measurement. However, you can at least put a lower bound on the problem if you know the tire dimensions and estimate how far into the ground it might sink (and still be acceptable).
 

Baluncore

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Unfortunately, it isn't that simple.
Car tyres, designed for performance on hard surfaces, are more rigid than the balloon-like tractor tyres used for traction on a soft soil.

The area of the tractor tyre contact patch, multiplied by the tyre pressure, will be very close to the load carried by the wheel. There may be variation in pressure across the contact patch due to imperfect directional elasticity. Deformation of the soil tends to be proportional to the local pressure of the tyre, which evens out the ground pressure, but the simple principle still holds.
Ground pressure is proportional to tyre pressure.
 

cjl

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I addressed that. On soft ground, if there's significant sinkage of the tire, the average contact patch pressure will be significantly below the tire's internal pressure.
 

Baluncore

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On soft ground, if there's significant sinkage of the tire, the average contact patch pressure will be significantly below the tire's internal pressure.
Then what is supporting the mass of the vehicle?
 
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Thank you for the calculations, Baluncore. Reading that a few more times, I think I'll be able to understand how this works.
Luckily my soft ground is just a 60m pass through from the road to my forest. I can't do dual wheels in the forest - they will become too wide.
 

cjl

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Then what is supporting the mass of the vehicle?
What do you mean? The air pressure is, of course, but some of that air pressure is opposed by tension in the tire sidewall rather than ground pressure. It's basically the opposite of a low pressure car tire on a hard ground, where the sidewall supports some of the load and thus the ground pressure is well above tire pressure.
 

Baluncore

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What do you mean? The air pressure is, of course, but some of that air pressure is opposed by tension in the tire sidewall rather than ground pressure.
The weight carried on a wheel is entirely transferred through the contact patch.
The average pressure on the patch must be weight / area.
The area of the patch will increase until it is opposed by the vertical component of the tyre pressure.
Tension in the sidewall is irrelevant as it does not press on the ground.
 

cjl

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The tension does not press on the ground, but it absolutely can counter some of the tire's internal pressure. Think of this in the extreme - if the tire were filled to some ridiculous pressure (say, 1000 PSI), then it would basically act as a rigid wheel. On a soft ground surface, it will still sink in a bit, and the ground contact pressure will be far, far less than this 1000psi internal pressure. All the internal force that isn't being opposed by ground forces is instead opposed by sidewall tension, just like it would be if there was no load on the tire.
 

Baluncore

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The tension does not press on the ground, but it absolutely can counter some of the tire's internal pressure.
That is obvious. If hoop tension in the tyre did not contain the internal pressure, the tyre would burst.

All the internal force that isn't being opposed by ground forces is instead opposed by sidewall tension, just like it would be if there was no load on the tire.
Consider two tyres with different internal pressures. When run against each other, they will share a contact patch determined by the lower of the two pressures.
The ground must be able to bear the tyre pressure, or the vehicle will sink and be unable to proceed. The tread on a tractor tyre is designed to pump fluid soil or mud on the surface outwards to expose the more solid ground below, where a contact patch may develop to provide traction.

Higher pressure tyres on hard surfaces have smaller contact patches with less sidewall deformation, which allows higher speeds without overheating of the tyre.
Tractors need lower tyre pressures with greater contact patches to avoid the soil shear that prevents traction.

Traction on soft ground is aided by lowering tyre pressure. If the tyre does not deform, but the soil does, then the principles and advantages of the pneumatic tyre cannot be applied.
 

cjl

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The ground must be able to bear the tyre pressure, or the vehicle will sink and be unable to proceed. The tread on a tractor tyre is designed to pump fluid soil or mud on the surface outwards to expose the more solid ground below, where a contact patch may develop to provide traction.
No, the ground must be able to bear enough pressure to support the vehicle. This need not be the tire pressure - it can be well below it, so long as it doesn't result in excessive sinkage of the tire. A very large tire helps here, because it allows the ground contact pressure to be very low without much sinkage because there is a large amount of tire contact with the ground.
 

Baluncore

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No, the ground must be able to bear enough pressure to support the vehicle.
I cannot understand why you are advocating large tyres with high pressure, that will compact the soil, leave deep wheel ruts, and quickly bog a tractor.

Large diameter tyres with high pressure will need heavy sidewalls to handle the hoop tension associated with the large radii of curvature and high pressure. Those tyres will be inflexible, heavy, and expensive. Lowering the pressure in those heavy tyres will destroy the more rigid side wall. Unfortunately your high pressure tractor tyre model is not beneficial to tractors in the real world.

If you over-inflate a tractor tyre, the tyre will sink into the soil only until the contact area between the soil and the rubber supports the tractor on the soil; the tyre will not deform. So high tyre pressure really guarantees the maximum possible ground pressure and soil deformation. At the same time it guarantees the minimum possible contact patch area. But traction is limited by shear of the soil, so traction requires the largest possible contact patch in order to reduce the horizontal shear stress in the soil below the driving wheels. So high tyre pressure denies benefit from the major advantages of pneumatic tractor tyres. On the other hand, low tyre pressure limits the maximum ground pressure to the tyre pressure, and allows the greatest contact patch.

The greatest revolution in tractors came when steel wheels that destroyed the soil, and often bogged the tractor, were replaced by low pressure pneumatic tyres that do not compact the soil and can provide a significantly greater draw bar force.
 

cjl

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I'm not advocating large tires with high pressure. I'm stating that with a soft ground, ground pressure can (and frequently will) be below tire pressure.
 

Baluncore

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I'm not advocating large tires with high pressure. I'm stating that with a soft ground, ground pressure can (and frequently will) be below tire pressure.
Are you saying that;
1. The average ground pressure of the contact patch will frequently be below the tyre pressure?

Or that;
2. The average will be made up of local areas with ground pressure higher than tyre pressure, and local areas with ground pressure lower than the tyre pressure?

Or that;
3. The contact patch will have a narrow periphery where ground pressure is tapered from tyre pressure to zero ground contact?

Or maybe something else?
 
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  1. If the contact patch is planar, then the ground load equals the pressure
  2. If the tire surface is convex then part of the pressure locally is due to the surface tension of the rubber tire . That local pressure equals the surface tension divided by some "average" curvature I believe.
  3. So if the tire patch is convex the internal pressure will be higher than the average pressure exerted by the ground.
Usually the contact patch is pretty flat.

I looked it up (see section 3.3.3

 
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