I'd be very interested to know how you got those numbers.
I'd also be interested to know why you are busting my balls!
Heath
"new" ZF - Bad Vibration at idle - Very Concerned
- Thread starter 84TD
- Start date
I'd be very interested to know how you got those numbers.
I'd also be interested to know why you are busting my balls!
Heath
can of worms
Since I did not take picture when I spun my 6.9/t19 flywheel on a crank balancer, I mounted it up to a wheel balancer.n I know its not a as sensitive as a engine balancer, but it will show an externally balanced flywheel, and I have easy access to the wheel balancer. The results were 1.5 ounces at 14.1 diameter. See pictures. So 1.5 oz probably does not seem like much compared to a late model 302 with 50oz in. of imbalance. If you sharp you noticed the in. behind the oz. What with the in. Well 50oz in means the 302 needs additional balance on the flywheel and damper of 50oz 1inch away from the center line of the crank or 25 oz 2 in away from the crank or 12.5 oz 4in away from the centerline of the crank and so on. How does this apply to that 1.5 oz on the 6.9 flywheel. Well that 1.5 oz is needed 7.05 inches away from the center of the flywheel to balance it. That means this flywheel has an imbalance of 1.5oz x 7 in or 10.5oz in. I would imagine the damper is similarly off balance. Assuming the imbalance is divided between the flywheel and damper the 6.9 would be 21ozin externally balanced. So we know now that the 6.9 flywheel is not 0 balanced. The spot the balancer wanted to add weight to the 6.9 flywheel was on the same side the flywheel was milled. 12 o’clock position in the picture of the flywheel. Unfortunately, I do not have a 7.3 flywheel to spin, but the imbalance should even be greater to account for the larger pistons.
How I came up with my previous estimate. I took a guess at the size of the trapezoidal boss. I assumed average width of 2”, 6” tall and .5 height. So 2inx6inx.5n=6ci. Machinery hand book states cast iron is .25lb/ci. 6ci.*.25lb/ci=1.5lb. The 3 holes are .5dia .75 deep. There is either a 118 or 135 degree cone on the bottom but we will assume it is flat at ¾ inch giving more volume removed. Area =pi x radius x radius. A=3.14x.25inx.25in. A=.196sqin. Volume is area time height. V=.196 sqin x .75in. V=.147sqin. Multiply time three each hole .441 sqin. .441 times 0.25equals .110 lbs.
So the boss weighs 1.5 lbs and the holes weigh .110 lbs. Or 24oz and 1.76oz..
Now if you read what was above you should be saying the center of mass of the 24oz bar is closer to the center of the flywheel then the 1.8 oz holes for the holes, and you would be correct. Also I think 1.5 lb is high for the boss after looking at the picture more closely and the 6.9 flywheel. Looking at the pictures the boss looks like 2 inches at the wide end 1 inch at the narrow 5 inches long and .25 height. I used the three ½ diameter holes to estimate the wide end of the boss. Long story short. 0.5(2in+1in)X5in X 0.25in X 0.25lb=.468lbs or 7.48oz.
I would say the center of the three holes are 7 inches from the center of flywheel.
The trapezoid boss is tougher. I cheated and used a cad program and descriptive geometry to find its center. I came up with 4.3 as the center of mass for the trapezoid.
The three holes 1.76oz X7 in= 12.32ozin.
The raised trapezoid boss 7.48 oz x 4.3in= 32.16 ozin.
32.16ozin – 12.32ozin=19.8 ozin of imbalance on a 7.3 flywheel in post 18.
note these are only estimates as I do not have an accessible 7.3 flywheel to measure. Thought about pulling the cover on mine to measure, but not really worth it.
This does not mean that you can feel the difference of imbalance from a 6.9 to 7.3, bearing wear over 100,000 miles will tell a different story.
Ford does not recommend the interchanging of 6.9, 7.3 and 7.3 turbo flywheels or dampers, must be a reason.
Just as a check, I calculated the weight of the milled out area of the 6.9 flywheel to be 4.93 oz with a center of 2.6 and came up with 12.818 ozin. Trust me on this calculations for area of circle segment is a bit long to put here. There are also a couple of shallow holes drilled on the outside on the side of the milled area. They would deduct 2.7ozin from the 12.818 to make 10.1ozin. 10.1 is pretty close to the measured 10.5 ozin of the 6.9 flywheel. In picture 3 you can see the milled area of the 6.9 flywheel.
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Since I did not take picture when I spun my 6.9/t19 flywheel on a crank balancer, I mounted it up to a wheel balancer.n I know its not a as sensitive as a engine balancer, but it will show an externally balanced flywheel, and I have easy access to the wheel balancer. The results were 1.5 ounces at 14.1 diameter. See pictures. So 1.5 oz probably does not seem like much compared to a late model 302 with 50oz in. of imbalance. If you sharp you noticed the in. behind the oz. What with the in. Well 50oz in means the 302 needs additional balance on the flywheel and damper of 50oz 1inch away from the center line of the crank or 25 oz 2 in away from the crank or 12.5 oz 4in away from the centerline of the crank and so on. How does this apply to that 1.5 oz on the 6.9 flywheel. Well that 1.5 oz is needed 7.05 inches away from the center of the flywheel to balance it. That means this flywheel has an imbalance of 1.5oz x 7 in or 10.5oz in. I would imagine the damper is similarly off balance. Assuming the imbalance is divided between the flywheel and damper the 6.9 would be 21ozin externally balanced. So we know now that the 6.9 flywheel is not 0 balanced. The spot the balancer wanted to add weight to the 6.9 flywheel was on the same side the flywheel was milled. 12 o’clock position in the picture of the flywheel. Unfortunately, I do not have a 7.3 flywheel to spin, but the imbalance should even be greater to account for the larger pistons.
How I came up with my previous estimate. I took a guess at the size of the trapezoidal boss. I assumed average width of 2”, 6” tall and .5 height. So 2inx6inx.5n=6ci. Machinery hand book states cast iron is .25lb/ci. 6ci.*.25lb/ci=1.5lb. The 3 holes are .5dia .75 deep. There is either a 118 or 135 degree cone on the bottom but we will assume it is flat at ¾ inch giving more volume removed. Area =pi x radius x radius. A=3.14x.25inx.25in. A=.196sqin. Volume is area time height. V=.196 sqin x .75in. V=.147sqin. Multiply time three each hole .441 sqin. .441 times 0.25equals .110 lbs.
So the boss weighs 1.5 lbs and the holes weigh .110 lbs. Or 24oz and 1.76oz..
Now if you read what was above you should be saying the center of mass of the 24oz bar is closer to the center of the flywheel then the 1.8 oz holes for the holes, and you would be correct. Also I think 1.5 lb is high for the boss after looking at the picture more closely and the 6.9 flywheel. Looking at the pictures the boss looks like 2 inches at the wide end 1 inch at the narrow 5 inches long and .25 height. I used the three ½ diameter holes to estimate the wide end of the boss. Long story short. 0.5(2in+1in)X5in X 0.25in X 0.25lb=.468lbs or 7.48oz.
I would say the center of the three holes are 7 inches from the center of flywheel.
The trapezoid boss is tougher. I cheated and used a cad program and descriptive geometry to find its center. I came up with 4.3 as the center of mass for the trapezoid.
The three holes 1.76oz X7 in= 12.32ozin.
The raised trapezoid boss 7.48 oz x 4.3in= 32.16 ozin.
32.16ozin – 12.32ozin=19.8 ozin of imbalance on a 7.3 flywheel in post 18.
note these are only estimates as I do not have an accessible 7.3 flywheel to measure. Thought about pulling the cover on mine to measure, but not really worth it.
This does not mean that you can feel the difference of imbalance from a 6.9 to 7.3, bearing wear over 100,000 miles will tell a different story.
Ford does not recommend the interchanging of 6.9, 7.3 and 7.3 turbo flywheels or dampers, must be a reason.
Just as a check, I calculated the weight of the milled out area of the 6.9 flywheel to be 4.93 oz with a center of 2.6 and came up with 12.818 ozin. Trust me on this calculations for area of circle segment is a bit long to put here. There are also a couple of shallow holes drilled on the outside on the side of the milled area. They would deduct 2.7ozin from the 12.818 to make 10.1ozin. 10.1 is pretty close to the measured 10.5 ozin of the 6.9 flywheel. In picture 3 you can see the milled area of the 6.9 flywheel.
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