# Wheel ground pressure vs wheel size

• petterg
In summary, the tractor needs a wider tyre to get similar ground pressure as the car tire in order to avoid leaving deep tracks. The available wheel widths are 520mm, 600mm, 650mm and 700mm, with a radius of 801mm required.
petterg
TL;DR 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?

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?

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?

petterg said:
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.

petterg said:
For these widths, which radius is required?

What Larger Standard Tire Diameters are available?

some alternatives:
620/60-42: 620mm wide, 967mm radius (this may not have enough space)

anorlunda said:
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?

petterg said:
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.

petterg said:
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.

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?

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.

petterg said:
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 into 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.

BvU
Baluncore said:
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).

cjl said:
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.

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.

cjl said:
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?

Tom.G
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.

Baluncore said:
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.

cjl said:
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.

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.

cjl said:
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.

cjl said:
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.

Baluncore said:
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.

cjl said:
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.

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.

cjl said:
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?

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

https://en.wikipedia.org/wiki/Surface_tension#Surface_curvature_and_pressure

Last edited:
Baluncore said:
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?
1, in cases where the ground is soft and deformable.

hutchphd said:
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

https://en.wikipedia.org/wiki/Surface_tension#Surface_curvature_and_pressure
Are we talking street tires on road surfaces here? For normal, passenger car radial tires, the contact pressure is usually well above the internal tire pressure, sometimes by as much as a factor of 2 at the full rated tire load. In general, contact pressure and internal tire pressure will be more similar under lower load and diverge under high load. They'll also be more similar for tires with tall sidewalls, and at higher tire pressure. Truck tires frequently will see contact patch pressures more similar to the internal pressure due to these reasons, as well as differences in the way they're constructed.

cjl said:
Are we talking street tires on road surfaces here? For normal, passenger car radial tires, the contact pressure is usually well above the internal tire pressure, sometimes by as much as a factor of 2 at the full rated tire load. In general, contact pressure and internal tire pressure will be more similar under lower load and diverge under high load. They'll also be more similar for tires with tall sidewalls, and at higher tire pressure. Truck tires frequently will see contact patch pressures more similar to the internal pressure due to these reasons, as well as differences in the way they're constructed.

My surface tension argument ignores any lateral stiffness of the tire material which I'm sure has some nontrivial effect. I am not a tire expert and I do not know how much. A steel radial belt is pretty stiff.

I was mostly pointing out that the "its just the interior pressure" argument fails in even the simplest case.

Can you point to some of the data you mention?... I might learn something.

It's irritatingly hard to find that data, unfortunately. You can kind of get a feel for some of the trends in this document, particularly figures 4.3-4.7, but I'm still looking for the passenger vehicle data I mentioned earlier (the linked document is truck tires). You can certainly see at least from the truck data that contact pressure and tire pressure, while related, are certainly not equal, but truck tires don't deviate as far from that expectation as passenger tires, due to the large sidewall, high load, and high inflation pressures.

Thanks. That is some data. I remember reading a paper about tractor tires contact when this post initiated. Can't seem to find it but will include it if I do... it was interesting. I have always been slightly awed by the durability and strength of tires.

I refer you to; Mechanics of pneumatic tires. NBS monograph 122. 1971.
By Gehman, S. D.; Ludema, K. C.; Backer, Stanley; Takeyama, T.; Matsui, J.; Clark, Samuel K.; Gough, V. E.; Walter, Joseph D.; Schallamach, A.; van Eldik Thieme, H. C. A.; Pacejka, H. B.

I quote Page 472;

In this section attention is directed first to the normal pressure distribution components caused by contact of the tire with some other surface. As a basic primary concept, one might state that
p = po+ f (Tire structural characteristics, tire driving or braking torque, tire side forces, tire velocity, etc.) (5.9)
where p is the vertical pressure component at any point, po is the inflation pressure of the tire and f is some general functional relationship which insofar as is now known is extremely complicated, and can best be described in a qualitative sense.
In eq (5.9), we postulate that the net pressure distribution at any point depends primarily upon the inflation pressure, and there is considerable experimental evidence to indicate that this is indeed true.

I couldn't get past p164 on your link (?) so I recommend this one:

https://archive.org/details/mechanicsofpneum122gehm/page/472
Thanks, it is a very interesting reference (I think I have defined myself as a nerd by that comment).
Notice the immediately subsequent equation eq.5.10, which mirrors what I was trying to say in comment #25 above!

hutchphd said:
Notice the immediately subsequent equation eq.5.10, which mirrors what I was trying to say in comment #25 above!
Eq.5.10 is referring specifically to where a tyre is tested against a rotating cylindrical roller. I agree that any test against a convex surface will give higher roller pressure results, which is the point of eq.5.10. But a road or an agricultural field is flat, not tightly curved.

Getting back to the OP.
petterg said:
I need to drive a tractor across and area of soft ground without leaving deep tracks.
I am talking about agricultural tractor tyres for use on soft ground. Internal pressures between about 12 and 25 psi, with large flat contact patches that do not leave ruts. Tractor tyres have a very flexible sidewall.

cjl said:
1, in cases where the ground is soft and deformable.
Under that situation the soil flows below the traction tyre to balance the ground pressure against the contact patch. Examine the track left by an agricultural tractor and you will see an equally compacted soil, with depressions where the tread lugs displaced soil. Soil flow is encouraged by horizontal soil shear due to traction.

The sidewall of a tractor tyre is folded with a short radius of curvature at the sides of the contact patch, which clearly precludes surface tension as a dominant factor. The front and rear of the contact patch have a tyre to ground separation angle that indicates surface tension will taper ground pressure from inflation pressure to zero at the entry and exit only.

Baluncore said:
Eq.5.10 is referring specifically to where a tyre is tested against a rotating cylindrical roller. I agree that any test against a convex surface will give higher roller pressure results, which is the point of eq.5.10. But a road or an agricultural field is flat, not tightly curved.
Unless it is mud...or other very viscous fluid. The fact that the resulting track is flat does not necessarily indicate that the instantaneous contact patch is flat. Mostly I was responding to a categorical statement about pressure and the contact patch. I did give a similar caveat in #25...
Clearly the biggest deviation in the real world comes from the rigidity of a tire and that analysis is well beyond my pay grade.

hutchphd said:
Mostly I was responding to a categorical statement about pressure and the contact patch. I did give a similar caveat in #25...
Clearly the biggest deviation in the real world comes from the rigidity of a tire and that analysis is well beyond my pay grade.

hutchphd said:
I was mostly pointing out that the "its just the interior pressure" argument fails in even the simplest case.
I do not claim that it is only interior pressure. I stand by my earlier statement for tractor tyres.
Baluncore said:
The ground pressure of a tyre is simply the tyre pressure.

All the pneumatic tyre references I have seen, and all my experience with tractor tyres indicates that primarily, the ground pressure is the same as the internal pressure.

You can argue that it does not hold around the edges, or for special cases of curved ground, or free flowing mud, but in any rational analysis it must be the first principle of pneumatic tyres.

<h2>1. What is wheel ground pressure?</h2><p>Wheel ground pressure is the amount of force per unit area that a wheel applies to the ground. It is typically measured in pounds per square inch (psi) or kilograms per square centimeter (kg/cm²).</p><h2>2. How does wheel size affect ground pressure?</h2><p>Generally, larger wheels will distribute the weight of a vehicle or object over a larger area, resulting in lower ground pressure. Smaller wheels, on the other hand, concentrate the weight over a smaller area, resulting in higher ground pressure.</p><h2>3. What are the benefits of having lower ground pressure?</h2><p>Lower ground pressure can help reduce soil compaction and minimize damage to delicate surfaces, such as lawns or agricultural fields. It can also improve traction and stability on soft or uneven terrain.</p><h2>4. Are there any downsides to having lower ground pressure?</h2><p>While lower ground pressure can be beneficial in certain situations, it can also limit the weight-carrying capacity of a vehicle or equipment. It may also require larger and more expensive tires, which can increase maintenance costs.</p><h2>5. How can wheel ground pressure be calculated?</h2><p>Wheel ground pressure can be calculated by dividing the weight of a vehicle or object by the total area of all the wheels in contact with the ground. This can be done using a formula or by using specialized equipment, such as a ground pressure gauge.</p>

## 1. What is wheel ground pressure?

Wheel ground pressure is the amount of force per unit area that a wheel applies to the ground. It is typically measured in pounds per square inch (psi) or kilograms per square centimeter (kg/cm²).

## 2. How does wheel size affect ground pressure?

Generally, larger wheels will distribute the weight of a vehicle or object over a larger area, resulting in lower ground pressure. Smaller wheels, on the other hand, concentrate the weight over a smaller area, resulting in higher ground pressure.

## 3. What are the benefits of having lower ground pressure?

Lower ground pressure can help reduce soil compaction and minimize damage to delicate surfaces, such as lawns or agricultural fields. It can also improve traction and stability on soft or uneven terrain.

## 4. Are there any downsides to having lower ground pressure?

While lower ground pressure can be beneficial in certain situations, it can also limit the weight-carrying capacity of a vehicle or equipment. It may also require larger and more expensive tires, which can increase maintenance costs.

## 5. How can wheel ground pressure be calculated?

Wheel ground pressure can be calculated by dividing the weight of a vehicle or object by the total area of all the wheels in contact with the ground. This can be done using a formula or by using specialized equipment, such as a ground pressure gauge.

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