Loader cylinder question

[tomstractorsandtoys]

On Deere 46a laoders there were 2 different lift cylinders. The first is 2 1/2 bore with 1 1/2 rod the second is 2 1/4 bore with 1 1/4 rod. Would the bigger cylinder be capable of lifting more? Tom

 

[Jim.ME]

Larger bore = more area for the pressure to act on, so yes, the 2-1/2'' cylinders should have more lifting force than the 2-1/4'' cylinders.

 

[VicS]

2.50 cyl compared to 2.25 cyl. Big is 4.91
sq in. Little is 3.97. So Little is 19
percent smaller. So that is quite a bit.

 

[Steve@Advance]

Here's a calculator, put in your bore and rod size, and an estimated PSI, it will give you a comparison.

You will find the retract force will be substantially less due to the rod displacement.
Cylinder Force Calculator

 

[Mike(NEOhio)]

Yes, lift more but a little slower.

 

[caterpillar guy]

Yup lift more and be more stable with the larger shaft. As for the comment about less retract strength. I've always been able to lift the front end off the ground with any loader I have used so how much more do I need?

 

[NY 986]

Force created is directly proportional to cylinder piston diameter.

 

[Indiana Ken]

(quoted from post at 06:31:41 10/09/23) Force created is directly proportional to cylinder piston diameter.

If I may comment: Yes the force increases with a larger diameter piston. However, the correct statement would be: "Force created is directly proportional to cylinder piston diameter squared".

You can also use the cross sectional area (see VicS) post, to calculate the increase in force.

 

[ss55]

In the 60s some loaders offered two sizes of cylinders to offer the same lifting capacity for tractors with very different maximum hydraulic pressures, say a two cylinder Deere vrs a New Generation Deere. The JD #45 is one example, it offered large diameter tie rod cylinders for low pressure systems and smaller diameter welded cylinders for the 2000 PSI systems on new generation tractors. The difference was much greater than 1/4 inch diameter though.

 

[Mike(NEOhio)]

The square of the radius, not diameter.

 

[Indiana Ken]

(quoted from post at 12:11:13 10/09/23) The square of the radius, not diameter.

Yes indeed, you can use the square of the radius, it will provide the same answer as using the square of the diameter.

Since a piston or bore is typically specified as a diameter, I prefer to go with the diameter. However, the radius will also work. Both methods will yield the same 19% reduction in force as calculated by "VicS" using the cross sectional area.