PDA

View Full Version : Converting rotational pull of a bolt to direct pull.

jeremy13
10-17-2012, 10:05 PM
I’m still looking for a way to pull a tone of loaded ammo. No make that close to 4 tons of loaded ammo. We have been playing with the standard RCBS collet type puller and it works quite well. And the change out to different calibers is only \$15.00. So expanding on this I want to put an air cylinder on the screw that pulls the collet tight. I tried getting a mesherment with a torque wrench. But my wrench only go’s down to 10lbs. and I’m less than that maybe a pull scale on the handle of the collet closer. It is only a handle 3” long. Question is once I get a value how do I convert it to pull force? The collet closer is a 3/8”X16tpi bolt. I know there is some mechanical advantage with the screw. So I need to add this to the equation. It only takes about 90deg turn of the collet closer from contact of the bullet tell tight enough to pull the bullet without slipping out. An air cylinder bigger than I need would be ok. I could fine tune the closing force with a regulator. I tried a 2” bore cylinder with a 3/4” rod at 100psi. It might have closed the collet just a little. Anyway it didn’t work. I could go with a lever arm if it is going to take a 6” bore cylinder for direct pull.

rohart
10-17-2012, 10:32 PM
If you want to work out how much force your closer, your bolt, is exerting when you hit the handle with a certain force, this is what you do. You treat the system as a lever. When you pull the handle, 3" long, round a full circle, you move it a distance of about 18" - 2 times pi times the radius. When that happens, the closer moves 1/16" - the tpi.

The ratio of that lever system is 18 divided by 1/16, or 288. So if you exert a force of 10lb at the end of the three inch handle, the closer is closing with a force of 2880lb, or 1 1/3 tons.

But this assumes a frictionless screw. Lets say that friction eats up one third of your effort. In this case a 10lb pull gives you a closing force of 1900lb.

I hope this is what you were after.

Abner
10-18-2012, 07:22 AM
http://www.spaenaur.com/pdf/sectionD/D48.pdf
From the chart...
400 in lbs torque = 5,000 lbs pressure = 1:12.5

edit; pressure on a 3" handle is 3x for conversion to in. lbs.
10 lbs. on a 3" handle is 30 in. lbs. x 12.5 = 375 lbs of force

BigJohnT
10-18-2012, 07:39 AM
I tried a 2” bore cylinder with a 3/4” rod at 100psi. It might have closed the collet just a little. Anyway it didn’t work. I could go with a lever arm if it is going to take a 6” bore cylinder for direct pull.

A 2" bore air cylinder with a 3/4" rod will not have much force at all retracting at 100psig due to the rod size. For an air cylinder Force = Pressure x Effective Area. Using a PHD CRS cylinder (http://www.phdinc.com/Product/?product=cylinders&series=CRS#Catalog) with a 2" bore as an example you get 304 pounds of force at extension and 260 pounds of force at retraction with the rod diameter of 3/4. Up that to a 2.5" bore CRS and you get 483 pounds at extension and 439 pounds at retraction. Quite a difference for just 1/2" bigger bore because of the area.

One trick I use when designing automation equipment to get more force into a small area is to stack two cylinders together. The bottom one a double rod cylinder and the top one a single rod cylinder.

John

rowbare
10-18-2012, 10:20 AM
Have you thought of using a stepper motor to tighten the screw? This thread describes using a stepper motor for a power drawbar setup: http://www.cnczone.com/forums/mechanical_calculations_engineering_design/154101-stepper-driven_power_drawbar.html

bob

Guido
10-18-2012, 11:10 AM
http://i126.photobucket.com/albums/p86/Guido_album/OTC4554-L.jpg

Ever use one of these when torqueing head bolts on a Texas puickup?

--G

rohart
10-18-2012, 01:03 PM
Yup - confession - forgot that little factor of twelve.

Well, it was the middle of the night, and it looked like no one was going to reply to Jeremy till todaqy, so I wiped the sleep out of my eyes... Just not enough.

Now if only the question had been in metric... My error would probably have been less, like, ten ?

Paul Alciatore
10-18-2012, 03:00 PM
Rohart's reply is right on and exactly how I do it. But do remember his warning about friction.

As for measuring torque values smaller than a regular torque wrench will allow, there are smaller versions. In the past I have purchased torque screwdrivers in various ranges, some quite low. McMaster has some. I would also think that larger torque wrenches are available too.

http://i126.photobucket.com/albums/p86/Guido_album/OTC4554-L.jpg

Ever use one of these when torqueing head bolts on a Texas puickup?

--G

This device appears to show an angle in degrees. How do you use it to find a torque value? Is there a table? If so, why would they do it that way instead of just calibrating it in ft-lbs or some other torque unit?

Paul Alciatore
10-18-2012, 03:15 PM
This power drawbar is super neat. Do check it out.

I have wondered if there was a simple way to power/automate the lock down screws on things like a RT for CNC operation. Perhaps a simpler version of this?

Have you thought of using a stepper motor to tighten the screw? This thread describes using a stepper motor for a power drawbar setup: http://www.cnczone.com/forums/mechanical_calculations_engineering_design/154101-stepper-driven_power_drawbar.html

bob

Black_Moons
10-18-2012, 04:03 PM
Rohart's reply is right on and exactly how I do it. But do remember his warning about friction.

This device appears to show an angle in degrees. How do you use it to find a torque value? Is there a table? If so, why would they do it that way instead of just calibrating it in ft-lbs or some other torque unit?

thats for 'torque past yield' fasteners I believe.
The idea being that your torque to X ft-lbs with a typical torque wrench, then turn them X degrees past that to actualy strech the bolt.. (And never use that bolt again if you reinstall the cylinder heads)

PS: rohart: That is a great method you came up with. Lots of websites out there use some calculation based on the *diamiter* of the bolt while not taking TPI into account at all!
Some people keep telling me this is correct.. Although none can exactly explain why the TPI has no effect, Clearly TPI provides a ratio for movement.
I am 100% convinced the TPI *has* to matter. Maybe those websites use the diamiter to lookup the TPI for a typical bolt (or whatever the TPI formual is for course bolts) but that can't be the same as for fine threaded bolts.

By reducing it to 'Move X, Z amount, results in moving Y, Z/R amount, Where R is the ratio of input force to output force, and ratio of distances moved' is very simple and can't really be wrong, based on my limited understanding of physics.

jeremy13
10-18-2012, 05:48 PM
OK I had some time to dig up a screw that I can replace the screw with handle. Now I can get a better reading on the torque required to pull the collet closed. I friend of mine has a up to 100 in-lbs. torque wrench well try it out to night and get a better reading.

Toolguy
10-18-2012, 06:14 PM
You might think about making a simple cam lock handle. That would be like a socket head cap screw (grade 8) with the head cut off, a handle with a cam on either side of the bolt and a 1/4" hardened steel dowel pin through the cam and bolt.
You would lift the handle to release the collet and push down the handle to tighten the collet. When the cam is at or just past center it will stay locked. You can adjust the tightness of the cam by screwing the bolt in or out.

uncle pete
10-18-2012, 07:07 PM
One more suggestion?
How about something set up like a air driven wedge type system, A bit like a morse taper collet only steeper so it's self releasing. Say 3 fingers inside a steep taper that are pulled or pushed up and down by an air cylinder on each finger. With 4 tons of ammunition to break down into it's components, your going to need something as automated as possible. And with that many cycles, the threads would wear fairly quick even if they were well lubricated. Acme would be far better of course.

Pete

jeremy13
10-18-2012, 09:42 PM
Friend came over with his very nice Snap-On torque wrench that went to 200 in-lb. And came up with 120 in-lb. or 10 ft-lb being about right. This is for what I’m considering to be the hardest of the rounds to pull apart. So 1800 lbs. of pull force is about right. This is just staggering to me never would have guessed that a screw can exert that much force with just a 90deg turn. Looks like it will take a 4” bore cylinder running at 150psi. To do the necessary clamping force. That will make things rather large. Fitting in all the cylinders will make this at least 30” wide. I’m thinking I would like to do 5 to 6 bullets at a time.

jeremy13
10-18-2012, 09:45 PM
This is what I'm using.
http://www.midwayusa.com/product/680804/rcbs-collet-bullet-puller
and the collets
http://http://www.midwayusa.com/product/128601/rcbs-collet-bullet-puller-collet-22-caliber-224-diameter?cm_vc=sugv1680804

kd4gij
10-19-2012, 12:17 AM
How about one of these http://www.harborfreight.com/air-tools/impact-wrenches/3-8-eighth-inch-drive-75-ft-lbs-torque-compact-air-impact-wrench-93100.html cheap and simple

Abner
10-19-2012, 08:55 AM
Hey guys,
There is a world of difference between 1.3 tons and 375 lbs. (which is correct). Can you spot the error? It's not friction.
It took me the better part of a day to get it.

BigJohnT
10-19-2012, 12:16 PM
Friend came over with his very nice Snap-On torque wrench that went to 200 in-lb. And came up with 120 in-lb. or 10 ft-lb being about right. This is for what I’m considering to be the hardest of the rounds to pull apart. So 1800 lbs. of pull force is about right. This is just staggering to me never would have guessed that a screw can exert that much force with just a 90deg turn. Looks like it will take a 4” bore cylinder running at 150psi. To do the necessary clamping force. That will make things rather large. Fitting in all the cylinders will make this at least 30” wide. I’m thinking I would like to do 5 to 6 bullets at a time.

You can stack smaller cylinders to get more force and you can use air over hydraulic for more force... and you can use boosters on smaller cylinders to get the air up to 300psig.

John

Bill736
10-19-2012, 01:30 PM
I once did an experiment to find the mechanical efficiency of my bench vise . The theoretical mechanical advantage was easy to measure; How far the jaws move when I turn the handle a measured distance ( assumes that the ball end of the handle is where the force is applied.) I then put my Dillon force gauge between the jaws, turned the vise handle until about 300 lb. of jaw force was indicated, and then attached a pull scale to the end of the vise handle to measure force. I avoided an error cause by static friction by moving the handle while observing the Dillon gauge. I was amazed to find that my vise was only about 20% efficient in converting handle force to jaw force. The remainder of the force was being eaten up by friction in the vise; mostly , I assume, in the screw. As I recall, I did oil the friction surfaces of the vise before my tests.

aostling
10-19-2012, 02:25 PM
I was amazed to find that my vise was only about 20% efficient in converting handle force to jaw force.

Your experiment confirms the value stated (apparently without a reference) in the Wiki article http://en.wikipedia.org/wiki/Screw_%28simple_machine%29: "Even well-lubricated jack screws have efficiencies of only 15% - 20%, the rest of the work applied in turning them is lost to friction."

The formula in that article states that the linear force F from a torque T on a screw of efficiency E

F = 2π TE/L where L = lead = 1/t.p.i.

Example: A torque of 100 in-lb on a screw with 12 threads per inch and 20% efficiency produces a force

F = (2)(3.14)(100)(.2)(12) = 1508 lb.

jeremy13
10-19-2012, 05:01 PM
Got out the nitrogen tank and put 150psi on the cylinder I have. It is metric so 50MM bore and a 20MM rod. I come up with 382 ft. pounds of pull force and still not enough. I took the torque wrench down to 90 in-lb. and it still pulled the bullets 80 in-lb. it would slip out. I’m figuring it both ways and I get the same two values. It’s looking like the 1800 ft-lb pull force might be the right way to figure it.

Abner
10-19-2012, 05:06 PM
It's all essentially about 2 different inclined planes the 6" diameter one and the .375" one.
16 tpi 1/16= .0625
Given; 3/8” bolt (.375”), 1 rev.= 1/16”(.0625”)
.375” diameter
Circumference of 3/8' bolt = 1.17”/.0625”= 18.72 x 30in lbs torque (10 lbs x 3” arm)
561.6 lbs minus friction and a reduction for actual thread diameter (from above cited chart 375 lbs.)
6” diameter
Circumference=18.84/.0625” = 301.44 X 10 (10 lbs applied at 3” radius)
3,014.4 lbs.
Abner

Rich Carlstedt
10-19-2012, 05:44 PM
If the leadscrew is a square thread, you will get axial force totally
If it is an Acme or V thread, then part of the force is radial

When putting Antiseize on a bolt thread, you only have half the objective. Place it under the head (washer face) as well to
reduce friction by 15 %

Rich