Center of Gravity Calculations and static tipping angles. Am I doing this right?

[LD1]

Bear with me, this will probably be a longer post than I intend.

The purpose of this exercise was to find the CG of my tractor in various configurations. So I broke out one of my college books on how to find the CG by raising one side up a given distance and seeing how much weight was transferred to the other axle. The higher you can go, the more accurate, which is part of my issue, since I can only go 12" comfortably. And any raising the rear any higher would cause me to have to elevate the FEL more giving another false reading.

So anyway, Following the formulas, I kept getting a really low CG height. Then it dawned on me that the formula works for cars cause the tires are the same diameter. So after browsing the net, I found a good calculator that will account for the different sizes. CG Height Calculator | Automotive Brake System Calculator | BRAKE POWER

So after plugging in all of the numbers:
Tractor, with FEL raised to 9.5", and bushhog with no weight on tailwheel: COG height was 22.3"
Tractor, with FEL raised to 9.5", and bushhog OFF, nothing else on 3PH...: COG height was 21.5"
Bare tractor, nothing on either end..................................................... : COG height was 23.8"

So not really much change in height. One thing that I dont understand....Having the bushhog low, gave a higher CG than with it off?

Now I know my weight measurements arent spot on. I was using a bottle jack with a 1000psi pressure gauge that has 20psi graduations. And the size of the jack, each PSI = 4lbs. But all three scenarios added 20psi to the front when raised, which is 80lbs. But I would say with the gauge, that could be a +/- 20lbs

But in any case, the CoG of all thee are close enough not to worry about, so going forward to CoG F/R location:

I want to calculate the static tipping angle of the tractor. The way I am doing that, which I believe is correct, seems to give awfully high numbers. But I know its static numbers and dont count for the dynamics of actually moving.

My rear tire width is identical to the wheelbase. 63.3". So using the stability triangle, we get an isosceles triangle with a 63.3" base and a 63.3" height. That couldnt have worked out any better, cause for each inch the height is shortened, the based is shortened by the same amount.

So.....Example 1 With FEL and Hog attached, the CoG is 21" forward of the rear axle and 22.3" high. Given the stability triangle, the tractor would have to be on an angle such that the COG would have to move 21.15" to the left or right to go outside the triangle. (not really the COG moving, but imagine a plumb bob attached at the COG point, and place the tractor on an angle. That plumb bob would have to move 21.15" at ground level) That angle would need to be 44.1

In example 2, with tractor and FEL, the CoG is 39.2" forward of the rear axle and 21.5" high. Given that it is much closer to the front axle, the stability triangle is much narrower up there and would only need the COG to move 12.05" before tip-over. That angle would be 29.24 degrees

Example 3 with bare tractor, the CoG is located 28.3" forward of the rear axle. About half way between Examples 1 and 2. And with a height of 23.8" A static tipping angle would be 36.34 degrees.

So here is the question: am I figuring the stability triangle and static tipping angle right? They seem like pretty high angles. I know at 22 degrees with the FEL and bushhog feels pretty uneasy, but according to this, I would need double in a static condition to upset. I know that at 22 degrees and wet grass, some sliding occurs. I know this dont count for the dynamics of moving, or me on the seat, or inertia of dropping a tire in a small hole.

All input is welcome. I know there are some pretty well versed members here in regards to the physics of this and math in general. So lets here it.

Edit: I also think it is important to note, that if my figuring is correct, while the FEL does lower the CoG, that dont mean it is more stable on a side slope.

 

[zzvyb6]

The very best way to determine the static incipient rollover angle is to tip it to that angle using a sidepull experiment. Either use a trailer to park the machine on, or pull it with a chain and a restraint bar. The complicating factor is the angle of the front axle where it hits any stops. Of course either a simulated or actual operator needs to be in place.

 

[dfkrug]

Well, doing the trigonometry, and then the real-world experiment are both valuable.

I read your description, LD, and perhaps you could clarify with a free-body diagram. Note
the wheelbase is the longitudinal distance between the wheels, and "wheel track" is the
width. Since you are interested in both to make the calc, and the latter to make the
tipover measurement, where you measure the width is quite variable. Tires are squishy and
wide, and you DO have to deal with the front axle pivot as ZZ mentions.

My $.002

 

[dfkrug]

Well, doing the trigonometry, and then the real-world experiment are both valuable.

I read your description, LD, and perhaps you could clarify with a free-body diagram. Note
the wheelbase is the longitudinal distance between the wheels, and "wheel track" is the
width. Since you are interested in both to make the calc, and the latter to make the
tipover measurement, where you measure the width is quite variable. Tires are squishy and
wide, and you DO have to deal with the front axle pivot as ZZ mentions.

My $.002

 

[LD1]

Dont really want to do that for risk of damage to the tractor.

And a side-pull and tilting of the front axle isnt the goal. A side pull (or jacking up one side) is going to put alot more stress on tires and axles than being on a hillside of the same degree.

When on a sideslope, the front axle pivot isnt going to hit the stops (until you start to go over that is.) because it will be on ground that is the same plane as the rear axle.

 

[LD1]

Well, doing the trigonometry, and then the real-world experiment are both valuable.

I read your description, LD, and perhaps you could clarify with a free-body diagram. Note
the wheelbase is the longitudinal distance between the wheels, and "wheel track" is the
width. Since you are interested in both to make the calc, and the latter to make the
tipover measurement, where you measure the width is quite variable. Tires are squishy and
wide, and you DO have to deal with the front axle pivot as ZZ mentions.

My $.002

Wheel track and rear wheel width for purposes of this experiment are the same. The actual track width is a bit wider. Almost 65" or a little better. but given the tires are somewhat rounded (think motorcycle tire) I just decided to use 63.3, cause it works out well with the stability triangle calculations, and also errors on the side of caution.

I'll see if I can get some drawings.

EDIT: and I am still not sure that the front axle pivot comes into play. At least not until a tip-over situation. Assuming that the ground is an even slope, it will be on the same angle as the rear.

 

[Sysop]

When on a sideslope, the front axle pivot isnt going to hit the stops (until you start to go over that is.) because it will be on ground that is the same plane as the rear axle.

come play on my slopes...

I have several areas that while traveling sideways, the pitch of the slope changes so much in a short area the front axle hits the stops both directions. These areas are steep enough you will slide sideways down the hill if conditions are not ideal. They have high pucker factor.

 

[bcp]

The Nebraska tractor testing organization did a study:

"Empirical equations to predict the tractor center of gravity."

Here on page 64 is the only place I found more than a reference. The math guys here will like this.
http://www.shin-norin.co.jp/english/pdf/VOL_40_NO3_SUMMER_2009.pdf


And here is another method:
AUTOMOBILE TESTING: METHOD FOR DETERMINATION OF CENTER OF GRAVITY OF AGRICULTURE TRACTORS
-------------------------------------------------------------------------
The "old farmer" methods I've heard:

Approximate side to side CG will be in line with the engine crankshaft.

Approximate front to back CG will be under the operator's feet.

Bruce

 

[mikehaugen]

One thing I think you are missing is that the the front axle pivot is not at ground level. It is actually an isosolese triangle with the vertex elevated. Not sure how to calculate it, but it will definitely change things.

 

[LD1]

One thing I think you are missing is that the the front axle pivot is not at ground level. It is actually an isosolese triangle with the vertex elevated. Not sure how to calculate it, but it will definitely change things.

That may be what I overlooked. Not sure how to correct for that though. Not sure if it will make things better or worse or if any different at all.

I know about everything that deals with the stability triangle deals with forklifts, and they always picture the triangle as being the baseline so to speak. It would seem that if the triangle were elevated, thus making it closer to the CoG, that would shorten the vertical componet of the triangle for calculating the angle to roll over, while the horizontal component stays the same, which would make the tipping angle greater?.....

Here are some sketched diagrams of how I did the math. I did notice I had an error in example one. Instead of 44.1 degrees, it is 43.5.



And I dont like the oldschool "approximate" CoG's.

While the vertical is probably close, the F/R split changes ALOT depending on weather there is a FEL, FEL loaded, 3PH attachment, etc. But it seems no matter how I loaded it, the CoG height didnt change much