Riddle me this: Shorter handle on single stage press

[onelight]

It just seems to me that is is going to require the same amount of force to resize a case and thus the same amount of force is being transferred through the the press to accomplish the same task
A shorter handle reduces the amount of force I can put on the press not increases it. Although with the shorter stroke I might increase the speed for a full stroke on the press. The short handle reduces my hand movement but changes nothing in the mechanical leverage designed into the press. I have a son that is an engineer for Boeing I get him draw me a picture

[M-Tecs]

It just seems to me that is is going to require the same amount of force to resize a case and thus the same amount of force is being transferred through the the press to accomplish the same task

It is correct that at the die the ram force is the same.

A shorter handle reduces the amount of force I can put on the press not increases it.

To get the same force at the die and ram with a shorter handle you need increase the pressure at the handle. This increased pressure is transferred to the pivot pin. In the case of a press there is so little difference it truly is a nonfactor. On something like a counter weight crane this becomes a real issue. The shorter the counterweight side is the more weight you need to balance. This means the pivot has more weight on it to balance.

[jmorris]

Very little load is on anything until the very last part of the stroke both in and out of the die. Then forces go very high, very quickly.

Maybe they figured if you had to ask, you shouldn’t be messing with it.?

Forster offers different length handles for their press and a number of others are adjustable from a Lee single stage all the way to the Dillon 1050.

[mdi]

If you used a lever with a fixed pivot point, and raised 50 lbs with a 6" lever, the only change if you used a 12" lever would be less force to move that 50 lbs. at the end of the lever. The same force would be on the pivot and the same force on the same length arm beyond the pivot point...

[M-Tecs]

If you used a lever with a fixed pivot point, and raised 50 lbs with a 6" lever, the only change if you used a 12" lever would be less force to move that 50 lbs. at the end of the lever. The same force would be on the pivot and the same force on the same length arm beyond the pivot point...

No sense in arguing about it when a test is so easy to do. Take a 12" ruler and lay it across a pivot point like a round pen. Place a quarter on each end so it balances. Now move one quarter half the distance to the pivot. Too get it to balance you have to add more weight to the the side you moved. This extra weight it transferred to the pivot.

[kenton]

The longer the handle the more leverage you will have and the shorter the less leverage you have. That is simple physics. The claim that there will more load on the pivot is correct in the case of the pivot pin on a press, however, this additional load is a non factor. It could be a factor on something like a kids teeter-totter. Example would be placing 100 pounds on both sides. It will balance and the force on the pivot will be 200 pounds plus the weight of the teeter bar. Now shorten ONE side by half. The shorter side will need an addition 100 pounds to balance so the load on the pivot is 300 pounds plus the weight of the teeter bar. If you keep shortening the one side the more weight you have to add to get to balance. At some point you may get enough weight that the pivot will fail. The extreme example would if one side of the teeter-totter is 10 feet long and the other side is much shorter. At 10 feet verse 1 foot the short side need 1,000 pounds to balance the 100 pounds. At 1 inch verse 10 feet you will need 12,000 pounds to balance the 100 pounds.

Since most presses now use a compound leverage system so this is even less of a factor. This compound leverage can produce thousands of pounds of force on the pivot. Adding an additional 20 or 30 pounds of downward force by choking up on the handle or using a short handle is a non factor in the case of a press.

That makes sense after I read the explanation twice. It also explains why portaging a tail heavy canoe wears out my shoulders faster.

[468]

[QUOTE=jmorris;4610589

Maybe they figured if you had to ask, you shouldn’t be messing with it...[/QUOTE]

My question to them, before they advised not to do it, was how much a new handle costs.

I’m not understanding your statement.

[onelight]

It is correct that at the die the ram force is the same.



To get the same force at the die and ram with a shorter handle you need increase the pressure at the handle. This increased pressure is transferred to the pivot pin. In the case of a press there is so little difference it truly is a nonfactor. On something like a counter weight crane this becomes a real issue. The shorter the counterweight side is the more weight you need to balance. This means the pivot has more weight on it to balance.

Ok this helps if you had a scale at the pivot on this example and the levers on both ends the same length with say 10lb on each end to balance your scale reads 20lb if you shorten one end so it requires 20lbs to balance the scale would read 30lbs so the load on the scale would have gone up 10lbs.
Aren’t you guys glad you weren’t my teachers in high school.

[468]

The longer the handle the more leverage you will have and the shorter the less leverage you have. That is simple physics. The claim that there will more load on the pivot is correct in the case of the pivot pin on a press, however, this additional load is a non factor. It could be a factor on something like a kids teeter-totter. Example would be placing 100 pounds on both sides. It will balance and the force on the pivot will be 200 pounds plus the weight of the teeter bar. Now shorten ONE side by half. The shorter side will need an addition 100 pounds to balance so the load on the pivot is 300 pounds plus the weight of the teeter bar. If you keep shortening the one side the more weight you have to add to get to balance. At some point you may get enough weight that the pivot will fail. The extreme example would if one side of the teeter-totter is 10 feet long and the other side is much shorter. At 10 feet verse 1 foot the short side need 1,000 pounds to balance the 100 pounds. At 1 inch verse 10 feet you will need 12,000 pounds to balance the 100 pounds.

Since most presses now use a compound leverage system so this is even less of a factor. This compound leverage can produce thousands of pounds of force on the pivot. Adding an additional 20 or 30 pounds of downward force by choking up on the handle or using a short handle is a non factor in the case of a press.

Great explanation!

[Rattlesnake Charlie]

Somebody failed physics class. In the 8th grade.

[44magLeo]

I can't dispute the claims M-Tecs.
Lets suppose the pivot point is designed to support 10,000 lbs. On the one end the lever is 10 ft long with 100 lbs on it. The other end is 10 ft with 100 lbs on. So load on the pivot point is weight of arms plus load on both arms. Just to keep it simple we'll work with just the load on the arms. Load is 200 lbs.
Let's move the load on one end in to 5 ft. This would need 200 lbs to balance the load thus giving a total load of 300 lbs. Lets even go a bit farther and move the one load into say one ft. That would need 1000 lbs to balance.
This is well with in the capacity of the pivot point.
Most of these heavy presses were designed so they have much more strength than needed to handle any possible load that even very heavy duty operations can be done. I can't think of anything much tougher to do than swaging bullets. Most people that do that often build their own presses for this. These presses are designed to make the task easier for the operator. They want presses that only need the 10 lbs to move the 1000 lb load not a 1000 lbs to move the 10 lb load.
RCBS first designed their presses to withstand swaging bullets and could do so for a very long time.
Resizing cartridges cases stress a press much less than swaging.
Sizing large cases like the 50 BMG is done on a press with about the same mechanical advantage as a regular press has but the an opening tall enough to allow the long case to be easily inserted into the press.
If the press is designed to withstand these severe loads and do so for several life times. Then doing regular loading will not overload anything even with a handle that is much shorter than the stock handle.
Leo

[M-Tecs]

44magLeo as I stated above the very minimal additional loading from a shorter handle in a loading press is a non-factor. While I have never put a load cell between the ram and die I do know you can generate at least a couple of thousand pounds of force due to the compounding linkage . The pivot pins are designed to withstand this level of pressure. Adding 20 or 30 more pounds to this level of pressure from a shorter handle is not an issue. The teeter-totter or balance cranes examples are just an easy way to help understand how a shorter handle will add a very minimal amount of additional loading to the pivot pin since additional force is required when using a shorter handle or chocking up on the handle. The output is the same regardless of handle length but handle length changes or even where you hold the handle changes the input force. This input force change is why the load on the pivot pin changes.

5 minutes later I get a call back from the Hornady rep advising that the engineers said DO NOT do this. You may damage the machine, and void warranty.

Technically the Hornady rep is correct that a shorter handle does introduce addition loading to the system, however, I would be really interested to know how an 20 or 30 pounds addition load to a system designed to deal with several thousand pounds of force is going to damage the press.

[David2011]

The problem I’m having with the teeter totter example is that the amount of work being done changes. In a loading press the amount of work remains fairly constant for a given operation such as sizing .44 Mag cases that were all loaded with the same powder charge and bullet. The work at the ram and therefore the pivot is a constant even though a 20” handle might require 20 pounds of force and a 10” handle would require 40 pounds of force to size the same cartridge.

[M-Tecs]

The problem I’m having with the teeter totter example is that the amount of work being done changes.

No. What changes is an increased load to account for the shorter point of leverage. The work (maintaining balance) output is the same.

The work at the ram and therefore the pivot is a constant even though a 20” handle might require 20 pounds of force and a 10” handle would require 40 pounds of force to size the same cartridge.

The additional 20 pounds of force has to go somewhere. As you stated the work at the ram is constant so the additional force is transferred to the first pivot pin.

[jmorris]

Technically the Hornady rep is correct that a shorter handle does introduce addition loading to the system, however, I would be really interested to know how an 20 or 30 pounds addition load to a system designed to deal with several thousand pounds of force is going to damage the press.

Just so I get this right, if I grab the handle between the ball and the pivot and stroke the press, effectively shorting it by 50%, I can break it?

[M-Tecs]

Just so I get this right, if I grab the handle between the ball and the pivot and stroke the press, effectively shorting it by 50%, I can break it?

No not even close. I don't know this for a fact but I have read that the compound press generate about 3,000 pounds of force at the ram. This load is also on the first pivot pin. Adding an additional 30 pounds would only increase load 1% max.

[rbuck351]

So if I turn the press upside down and shortened the handle this would lessen the pressure on the link pin? Or if I left the handle long so they would balance and just grabbed it shorter the pin load would be the same? Or if I took the press to space in 0 gravity the pin load would be the same short or long handle? If the only difference is the difference in the weight of the shorter handle we could add say a pound or so counter weight to the end of our shorter handle to get balance and the added weight of the counter balance would be the only difference in pin pressure. This is all nice in theory but a pound or so of different load is not even worth mentioning let alone thinking it could damage a press.

[M-Tecs]

So if I turn the press upside down and shortened the handle this would lessen the pressure on the link pin?

No turning the press upside down would not effect the force vectors in the press and the shorter handle still requires an increased load to do the same amount of work. Same for zero gravity not effecting the force vectors in the press other than the actual weight of the components.

If you are having a hard time visualizing it drawn yourself a diagram including the loads and force vectors.

[onelight]

The extra loading on the pivot is = to the extra pressure you feel on your hand when you move your hand to the center of the handle on your press

[rbuck351]

There is no weight in zero gravity and shortening one side of a fulcrum by half would double the amount of force required to do a given amount of work on the opposite side. No more no less. The only difference of force on the pin in a gravity area is the missing piece of the handle and it's distance from the fulcrum. Maybe an additional pound or two. Check it out with a torque wrench.
Lock the square drive in a vice and apply a given weight on the end. Then double the weight at mid point and see if the torque spec doesn't read the same.