Anonymous Poster
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- Sep 27, 2005
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Hi,
I got curious about where my tractor might tip over as expressed in degrees of slope. Naturally, trying to calculate this is impossible if the tractor is moving, at least if one is limited by the head I happen to have resting on my shoulders /w3tcompact/icons/smile.gif.
BUT...STILL... it seems like a static calculation could be of some usefulness...even if the amount of usefulness it might have is debatable.
It seemed to me that the tractor center of gravity could be conservatively placed between the top of the rear tires and the top of the wheel rims. My rear tires are filled with WW fluid to the top of the rims, and the front tires are foam filled. Actually the center of gravity might be below the tops of the rear wheels. If so, that would add to the stability of the tractor.
I made some measurements on my Kubota B2910, which has R4 tires. They came out as follows:
Tire height to top of tread: 37 inches
Height to top of rear rims: 26 inches
Width of tire to outside of tread contact patch: 52 inches
Width of tire tread to inside of contact patch: 29.5 inches
Since the tractor is for practical purpose symmetrical, the center of gravity should, for practical purposes, be centered between the rear tires. So the variables of most importance, is the vertical position of the center of gravity, and the place on the surface of the tire contact patch that is effectively supporting the tractor. Both of these are really variables and unknown to me. But I can guess the limits of possibilities.
The limits are the outside and inside edges of the tire contact patches, and the maximum and minimum range of likely heights of the center of gravity.
Using these limits it should be possible to calculate the best and worse case static tip over angles.
Again, let me emphasize STATIC, as once the tractor starts moving other factors come into play that can reduce the static angles considerably.
The worst case is having the center of gravity at the tops of the tires and the point at which the tractor is supported at the inside edges of the tires. For my tractor, IF this were the case [I am sure it is not] the calculated angle at which the tractor would tip over would be a slope of 21 degrees!
The other extreme would be the lowest center of gravity position and the widest support point at the outer edges of the tire contact points. This calculates out to a slope angle of 45 degrees!
The midpoint between the two extremes calculates out to be a slope of 33 degrees where the tractor would roll over.
Again, this would be without any movement. A rock or hole that a front wheel hits would lower these numbers considerably!
Here�fs a summary:
(Degrees)
Height <font color=blue> 1/2 Width</font color=blue> Tip Angle
37 inches <font color=blue> 14.34 inches</font color=blue> 21 Degrees
31.5 inches <font color=blue> 20.5 inches</font color=blue> 33 degrees
26 Inches <font color=blue> 26 inches </font color=blue> 45 degrees
1/2 the tread width is used for the calculation, because the point of tipping occurs when the center of gravity to beyond the point directly above the tire's point of contact.
Like I said, these numbers are interesting, but worthless, because of the other variables that come into play when we are using our tractors. Like the weight of the operator and how the operator himself changes the center of gravity of the tractor/operator combination. Or what's hanging on the back. Or, the height of the loader and what's inside it. Or..or...or...
Or the dynamic effects of the tractor moving and hitting a bump/hole.
Still, this exercise in futility has given me something, as I may be on a slope and be maneuvering at about a speed of 1 foot per minute, and feeling pretty unsure of myself.
I think I may feel a little more comfortable in an almost stopped situation now...
Prior to this, I was puckering at probably 15 degrees when stopped. Just had no feel for anything at all. Now I will pucker at 15 degrees and enjoy it!
I now at least have a little gut feeling about what the maximum limits MIGHT be...
This means something to me, as I can�ft bring myself to go out and traverse increasingly steep slopes until the tractor rolls over for educational purposes!
In another thread where I posted some pictures of slopes that were mowed that looked crazy to me, some others felt that they were doable mowing without too much concern.
My simple calculations support their observations.
A 20 degree cross slope still seems quite steep to me though. That is about the maximum on my property and I have not been across one yet.
For what it�fs worth...this exercise was of some interest to me!
<font color=red>Don't forget, these numbers mean nothing in the real world!</font color=red>
Any comments?
Bill in Pgh, PA
I got curious about where my tractor might tip over as expressed in degrees of slope. Naturally, trying to calculate this is impossible if the tractor is moving, at least if one is limited by the head I happen to have resting on my shoulders /w3tcompact/icons/smile.gif.
BUT...STILL... it seems like a static calculation could be of some usefulness...even if the amount of usefulness it might have is debatable.
It seemed to me that the tractor center of gravity could be conservatively placed between the top of the rear tires and the top of the wheel rims. My rear tires are filled with WW fluid to the top of the rims, and the front tires are foam filled. Actually the center of gravity might be below the tops of the rear wheels. If so, that would add to the stability of the tractor.
I made some measurements on my Kubota B2910, which has R4 tires. They came out as follows:
Tire height to top of tread: 37 inches
Height to top of rear rims: 26 inches
Width of tire to outside of tread contact patch: 52 inches
Width of tire tread to inside of contact patch: 29.5 inches
Since the tractor is for practical purpose symmetrical, the center of gravity should, for practical purposes, be centered between the rear tires. So the variables of most importance, is the vertical position of the center of gravity, and the place on the surface of the tire contact patch that is effectively supporting the tractor. Both of these are really variables and unknown to me. But I can guess the limits of possibilities.
The limits are the outside and inside edges of the tire contact patches, and the maximum and minimum range of likely heights of the center of gravity.
Using these limits it should be possible to calculate the best and worse case static tip over angles.
Again, let me emphasize STATIC, as once the tractor starts moving other factors come into play that can reduce the static angles considerably.
The worst case is having the center of gravity at the tops of the tires and the point at which the tractor is supported at the inside edges of the tires. For my tractor, IF this were the case [I am sure it is not] the calculated angle at which the tractor would tip over would be a slope of 21 degrees!
The other extreme would be the lowest center of gravity position and the widest support point at the outer edges of the tire contact points. This calculates out to a slope angle of 45 degrees!
The midpoint between the two extremes calculates out to be a slope of 33 degrees where the tractor would roll over.
Again, this would be without any movement. A rock or hole that a front wheel hits would lower these numbers considerably!
Here�fs a summary:
(Degrees)
Height <font color=blue> 1/2 Width</font color=blue> Tip Angle
37 inches <font color=blue> 14.34 inches</font color=blue> 21 Degrees
31.5 inches <font color=blue> 20.5 inches</font color=blue> 33 degrees
26 Inches <font color=blue> 26 inches </font color=blue> 45 degrees
1/2 the tread width is used for the calculation, because the point of tipping occurs when the center of gravity to beyond the point directly above the tire's point of contact.
Like I said, these numbers are interesting, but worthless, because of the other variables that come into play when we are using our tractors. Like the weight of the operator and how the operator himself changes the center of gravity of the tractor/operator combination. Or what's hanging on the back. Or, the height of the loader and what's inside it. Or..or...or...
Or the dynamic effects of the tractor moving and hitting a bump/hole.
Still, this exercise in futility has given me something, as I may be on a slope and be maneuvering at about a speed of 1 foot per minute, and feeling pretty unsure of myself.
I think I may feel a little more comfortable in an almost stopped situation now...
Prior to this, I was puckering at probably 15 degrees when stopped. Just had no feel for anything at all. Now I will pucker at 15 degrees and enjoy it!
I now at least have a little gut feeling about what the maximum limits MIGHT be...
This means something to me, as I can�ft bring myself to go out and traverse increasingly steep slopes until the tractor rolls over for educational purposes!
In another thread where I posted some pictures of slopes that were mowed that looked crazy to me, some others felt that they were doable mowing without too much concern.
My simple calculations support their observations.
A 20 degree cross slope still seems quite steep to me though. That is about the maximum on my property and I have not been across one yet.
For what it�fs worth...this exercise was of some interest to me!
<font color=red>Don't forget, these numbers mean nothing in the real world!</font color=red>
Any comments?
Bill in Pgh, PA