Will This Work

[bmwmike]

Not sure how to figure the % of lead for the tires I want to try?
I have a B7100 with turfs. Front 20.5x8-10 Rear 29x12.5-15.

Has anyone considered putting these on turf rim for more traction.

Front Titan Tru power
18x8.50-10
Rolling Circumference 56"
Overall Dia. 18.5"
Rim 7"

Rear Titan Trac loader
27x10.50-15
Rolling Cir. 82"
Overall Dia. 27.5"
Rim 8.5"

I know the rear rims are 1.5" less , but if I have to I can put in tubes to pop it out to the larger rim.

Can someone who khow how do the math and tell me if this combo will have the correct lead time, or how far off they would be.

Thanks
Mike

 

[lsmurphy]

-6.84%

Outside the 0-5% lead.

May not work.

Also you cannot mount a 10"w tire to a 12"w wheel. This will also change the RC of the rear tire

 

[lsmurphy]

62 x 1.464 = 90.768 divided by 88 = 1.0314 minus 1 x 100 = 3.14 %

well within the 0-5% lead recommended by Kubota


Above is the formula...........change the 62 and the 88 with your tires RC

 

[bmwmike]

Ok, How about this.

Front Titan Tru Power
18x8.50-10
Rolling cir. 56"
Rim 7"

Rear Goodyear Super Terra Grip
29x12.50-15
Rolling cir. 86"
Rim 10"

56X1.464=81.984 divided by 86=0.9533023 - 1= -0.0466977 x 100= -4.66977%

If I did that right it is within the -5%.

mike

 

[bmwmike]

I just reread your answer and I think you are saying that it should be a lead from 0 to 5% and I was reading as 0 to -5%.

How did you get -6.84%?

56x1.464=81.984 divided by 82= 0.9998048 - 1= -0.0001952 x 100= -0.01952

Which is real close to 0, and I see what you are saying about the rim size changing the rolling cir. but I should be able to make up for that by inflating or deflating the tires, Maybe? Any way this is just and Idea for useing my turf tire rims and getting more traction.

mike

 

[bmwmike]

Here is another combo that is a possibility.

Front Titan HD 2000
20x8.00-10
Rolling cir 57"
overall dia. 19.3"
Rim 6"
Tire width 8.2"
max load 1210 lbs.

Rear Titan Trac Loader
27x10.50-15
rolling cir. 82"
overall dia. 27.5"
tire width 11"
rim 8.5"
max load 2610 lbs

My rims are front 6" , rear 10" . so I would be putting the rear tire on a rim 1.5" larger than called for, even if this changes the rolling cir. by 1 " + or - it still falls in the 0 to 5% lead they call for .

57x1.464=83.448 divided by 82 = 1.0176585 - 1 = 0.0176585 x 100= 1.76585%

 

[lsmurphy]

Your current 12.5w tire should be mounted to a 10" wheel. You'll need a 8" wheel to mount a 10" tire.

 

[lsmurphy]

57x1.464=83.448 divided by 82 = 1.0176585 - 1 = 0.0176585 x 100= 1.76585%

This looks correct.

 

[tiredguy]

bmwmike,
first and foremost you need the rolling circumference of what was on it in order to figure out the correct "ratio" you need to be at. What you've come up with is front RC 56 divided by rear RC 85 equals .68. You have to be within 2% ( the closer the better ) in order to be within the tolerence to not cause you any problems.

Keep in mind that rolling circumference is determined by the tire being mounted on the correct rim width and that's where most people go awry making mistakes. If the rim is to narrow it will cause the tire to be larger in diameter significant enough to be way off the 2% mark as would to wide of rim for the tire to be smaller than what the specs list it to be.

Don't waste your time with other formulas unless you're a very well educated engineer with excellent math skills too. Putting a tire on a narrower or wider rim throws out what is shown for rolling circumference. The only way you can figure correctly what the actual numbers are is to mount the tires up install them on the tractor, them mark and measure the distance traveled.
Al