Tool Talk
General Category => General Discussion => Topic started by: Twilight Fenrir on August 05, 2014, 12:48:57 PM
-
Well, I went to go look at a giant antique bandsaw the other day, and the guy had a few other interesting things he was selling... A MASSIVE mechanical automotive floor jack (non-hydraulic), a fixed position winch, and this little arbor press... It's marked a 2 ton press... But, there's something missing on this, right? How can such a little thing put out two tons of force with just a rack and pinion?
Thanks,
Richard
(http://i137.photobucket.com/albums/q226/Midnight_Fenrir/Tool%20Pron/IMG_20140804_185159_zpsb09fac22.jpg)
-
Leverage.
-
>Leverage
Hmm..
If the rack gear is 1/2 inch across, and 2 tons is 4000 lbs, you need a lever 200 inches long with 10 lbs on it...
Or a 400 lb fellow with long arms.....
-
Rusty is on the right track......
It's all relative, the radius of the drive gear, the length of the handle, the weight on the handle end, etc.
I suspect the press rating is the maximum force that it can handle, rather than a measurement of its press power.
But - this is a simple machine, a lever. Instead of two lengths on either side of a pivot this has one length attached to a gear that transfers the force at a right angle down to the shaft. The gear is kind of the key. If the end of the handle moves 10" when the press moves 10" down the force pushing down on the handle would equal the force pressing down through the press, almost (a small amount of energy would be lost to friction and would be turned into heat, other energy could be lost bending components of the system.) The gear radius would then be the length of the input lever. The torque at the gear moves the shaft. The distance from the center of the gear to the point it applies force through the shaft is like one side of a see-saw, the other side is the point on the handle where force is applied at to the center of the gear. The center if the gear is the pivot point.
So how is the downward force applied to the handle amplified?
The travel distance of each of the points where the force is applied is needed to determine how much the force is multiplied, or rather to cslculate the ratio of input force to output force.
Moment is the force times distance. If we consider the effect of friction and energy lost flexing the bar to be negligible the input moment will equal the output moment.
Formulaicly stated the input force times the input length equals the output force times the output length.
Given an input arm length of 10" with a gear radius of 1" the ratio is 10 to 1.
A simple measurement using the picture and my finger as a ruler looks like the ratio of (handle to center of gear) to (center of gear to center of press shaft) is around 12 to 1.
Based on a 2 ton rating, and my inaccurate finger measurement, you should be able to apply 333-1/3 lbs of weight on the end of the handle safely. (Don't use this, I am guessing on the lengths.)
Measure the distance from:
A. The center of the gear to the point you pull down on the handle
and
B. The center of the gear to the center of the press shaft.
Alternatively, measure the distance the handle travels and the distance the press shaft travels. Remember to take the handle distance at the point you apply the force (about the middle of where you grab it.)
The ratio is handle distance to shaft distance.
If shaft distance is 1/2" and handle distance is 10", the ratio is (10) 20 to (1/2) 1. About 200 lbs max input then.
Long explanation for what Rusty almost said. Sorry for that. :embarrassed:
Chilly
-
We had one at a shop I worked at similar to the one in the photo. It's purpose was a Ujoint press, about the limit
unless you put a pipe on the handle. Ok for putting bearings on , sometimes wouldn't take them off.
Better than a hammer.
-
Rusty is on the right track......
It's all relative, the radius of the drive gear, the length of the handle, the weight on the handle end, etc.
I suspect the press rating is the maximum force that it can handle, rather than a measurement of its press power.
But - this is a simple machine, a lever. Instead of two lengths on either side of a pivot this has one length attached to a gear that transfers the force at a right angle down to the shaft. The gear is kind of the key. If the end of the handle moves 10" when the press moves 10" down the force pushing down on the handle would equal the force pressing down through the press, almost (a small amount of energy would be lost to friction and would be turned into heat, other energy could be lost bending components of the system.) The gear radius would then be the length of the input lever. The torque at the gear moves the shaft. The distance from the center of the gear to the point it applies force through the shaft is like one side of a see-saw, the other side is the point on the handle where force is applied at to the center of the gear. The center if the gear is the pivot point.
So how is the downward force applied to the handle amplified?
The travel distance of each of the points where the force is applied is needed to determine how much the force is multiplied, or rather to cslculate the ratio of input force to output force.
Moment is the force times distance. If we consider the effect of friction and energy lost flexing the bar to be negligible the input moment will equal the output moment.
Formulaicly stated the input force times the input length equals the output force times the output length.
Given an input arm length of 10" with a gear radius of 1" the ratio is 10 to 1.
A simple measurement using the picture and my finger as a ruler looks like the ratio of (handle to center of gear) to (center of gear to center of press shaft) is around 12 to 1.
Based on a 2 ton rating, and my inaccurate finger measurement, you should be able to apply 333-1/3 lbs of weight on the end of the handle safely. (Don't use this, I am guessing on the lengths.)
Measure the distance from:
A. The center of the gear to the point you pull down on the handle
and
B. The center of the gear to the center of the press shaft.
Alternatively, measure the distance the handle travels and the distance the press shaft travels. Remember to take the handle distance at the point you apply the force (about the middle of where you grab it.)
The ratio is handle distance to shaft distance.
If shaft distance is 1/2" and handle distance is 10", the ratio is (10) 20 to (1/2) 1. About 200 lbs max input then.
Long explanation for what Rusty almost said. Sorry for that. :embarrassed:
Chilly
My goodness! Thank you for the very thorough answer! I suppose it really is surprising what kind of force a simple machine can generate. It didn't feel like it could put a hundred pounds into something, when I was fiddling with it on the floor in that garage... But, I wasn't about to stick my finger under it and test that theory :P
I didn't take any measurements at the time, nor did I take any other pictures... I thought for sure it was missing some major component, and wouldn't be worth my time...
We had one at a shop I worked at similar to the one in the photo. It's purpose was a Ujoint press, about the limit
unless you put a pipe on the handle. Ok for putting bearings on , sometimes wouldn't take them off.
Better than a hammer.
Hmm, well, I don't often have applications where I NEED a press, but I've had to change bearings now and then on my cars, and they were a BEAR, especially on my '66 Toronado... but, those were like... 4" in diameter, pressed on to a honking chunk of steel :P Doubt this little press would have helped me much on that one...
Still, depending on what he wants for it, I might pick it up then, could be useful n.n
-
GM truck ball joints were like that, I had a set laugh at a 6 ton hydraulic press once...
Then they laughed at a 12 ton....
30 ton press across town and a sledge hammer to get them out of the a-arms....
Sure...any home mechanic can change them ;P
-
where are the pics of the bandsaw and jack??
Skip
-
Don't have one of the jack... But here's the saw...
(http://i137.photobucket.com/albums/q226/Midnight_Fenrir/Tool%20Pron/IMG_20140804_184145_zps8ed4bea3.jpg)
And the winch:
(http://i137.photobucket.com/albums/q226/Midnight_Fenrir/Tool%20Pron/IMG_20140804_184924_zps8cab39a9.jpg)
-
did you identify the band saw ?
are you getting it ?
it would be a nice save.
Frank
-
did you identify the band saw ?
are you getting it ?
it would be a nice save.
Frank
Well, there are no markings on the saw... But, over at OWWM, we believe we've got it pinned down to be a Frank & Co. No 2 bandsaw. Weighing in at 800 lbs, with 30 inch wheels a 12 resaw, and 30" throat clearing... Way overkill for what I need... But, yes, I am picking it up Saturday morning, assuming we can get it in my truck :P
(http://www.owwm.org/download/file.php?id=52067&sid=4cb292c1a412bf79e3b843e10d38b30a&mode=view)
-
Looks to be close to 8' tall. Very nice machine. I'm glad you are saving it.
Too many are scrapped.
Chilly
-
Also need to consider the square inch of press... 2 tons over 1 Square inch isn't hard, considering a high heal shoe spike can apply ~1 ton pressure over 1/4" sq.....
-
A friend I'd known since he was in diapers once asked over the phone if I ever saw big bandsaws for sale. He and a buddy were in New Mexico (I'm in north-central Idaho) and were wanting to bandsaw decorative ends on Spanish-style work. I said "No" but within two weeks I checked out the offerings at an upcoming auction, which included a Eureka #1 from the now-gone Vessey Sawmill, in Kooskia, ID. They gave me a $1,500 limit. I bo't the saw, several spare bands and a spot-welder made specifically for bandsaw blades, plus several buckets of bolts and "stuff", all for $320, if memory serves. They about fell over themselves getting up here and totally reimbursed me for everything (even "my" stuff) at that price. We used a backhoe to lay that huge sucker down on a flatbed trailer and carefully fit the small stuff in and around it — and away they went. As I always say, "The impossible just takes a little longer," which in this case didn't take long at all. Thanks for the memories, the story and all.
-
A friend I'd known since he was in diapers once asked over the phone if I ever saw big bandsaws for sale. He and a buddy were in New Mexico (I'm in north-central Idaho) and were wanting to bandsaw decorative ends on Spanish-style work. I said "No" but within two weeks I checked out the offerings at an upcoming auction, which included a Eureka #1 from the now-gone Vessey Sawmill, in Kooskia, ID. They gave me a $1,500 limit. I bo't the saw, several spare bands and a spot-welder made specifically for bandsaw blades, plus several buckets of bolts and "stuff", all for $320, if memory serves. They about fell over themselves getting up here and totally reimbursed me for everything (even "my" stuff) at that price. We used a backhoe to lay that huge sucker down on a flatbed trailer and carefully fit the small stuff in and around it — and away they went. As I always say, "The impossible just takes a little longer," which in this case didn't take long at all. Thanks for the memories, the story and all.
Hahaha, if I could find someone who wanted to pay $1000 for it, I'd sell it in a heartbeat :P but that's pretty neat!
-
Also need to consider the square inch of press... 2 tons over 1 Square inch isn't hard, considering a high heal shoe spike can apply ~1 ton pressure over 1/4" sq.....
In this case it is quite a lot.
I don't understand the high heel analogy, could you help me? Is the wearer very large, or are you referring to stomping or something like that?
-
Also need to consider the square inch of press... 2 tons over 1 Square inch isn't hard, considering a high heal shoe spike can apply ~1 ton pressure over 1/4" sq.....
In this case it is quite a lot.
I don't understand the high heel analogy, could you help me? Is the wearer very large, or are you referring to stomping or something like that?
Well, that's a bit of an exaggeration, but right in principle...
If a person weighs 200 lbs is standing on 1/4" spike, that's 800 psi. Doesn't quite work out that way, because SOME weight is on the front of the shoe, and there are two heels. If you narrowed the heel down to a real spike, pointy and all, could clear 2k psi easily, I suspect.
-
Also need to consider the square inch of press... 2 tons over 1 Square inch isn't hard, considering a high heal shoe spike can apply ~1 ton pressure over 1/4" sq.....
In this case it is quite a lot.
I don't understand the high heel analogy, could you help me? Is the wearer very large, or are you referring to stomping or something like that?
Well, that's a bit of an exaggeration, but right in principle...
If a person weighs 200 lbs is standing on 1/4" spike, that's 800 psi. Doesn't quite work out that way, because SOME weight is on the front of the shoe, and there are two heels. If you narrowed the heel down to a real spike, pointy and all, could clear 2k psi easily, I suspect.
Yes, but to show the "point", that you Also need to consider the "square inch of press" are you pressing more than 1" square? or less
1" sq. is 16 1/4" sq. ( 4X4), so 16 X 200 is 3200#, (granted she is standing on 1 heel.) something to think about...
-
Also need to consider the square inch of press... 2 tons over 1 Square inch isn't hard, considering a high heal shoe spike can apply ~1 ton pressure over 1/4" sq.....
In this case it is quite a lot.
I don't understand the high heel analogy, could you help me? Is the wearer very large, or are you referring to stomping or something like that?
Well, that's a bit of an exaggeration, but right in principle...
If a person weighs 200 lbs is standing on 1/4" spike, that's 800 psi. Doesn't quite work out that way, because SOME weight is on the front of the shoe, and there are two heels. If you narrowed the heel down to a real spike, pointy and all, could clear 2k psi easily, I suspect.
Yes, but to show the "point", that you Also need to consider the "square inch of press" are you pressing more than 1" square? or less
1" sq. is 16 1/4" sq. ( 4X4), so 16 X 200 is 3200#, (granted she is standing on 1 heel.) something to think about...
.... Right you are... I never did do very well in geometry XD Could throw algebra at me all day long, but for some reason, geometry just never stuck...
-
Also need to consider the square inch of press... 2 tons over 1 Square inch isn't hard, considering a high heal shoe spike can apply ~1 ton pressure over 1/4" sq.....
In this case it is quite a lot.
I don't understand the high heel analogy, could you help me? Is the wearer very large, or are you referring to stomping or something like that?
Well, that's a bit of an exaggeration, but right in principle...
If a person weighs 200 lbs is standing on 1/4" spike, that's 800 psi. Doesn't quite work out that way, because SOME weight is on the front of the shoe, and there are two heels. If you narrowed the heel down to a real spike, pointy and all, could clear 2k psi easily, I suspect.
Yes, but to show the "point", that you Also need to consider the "square inch of press" are you pressing more than 1" square? or less
1" sq. is 16 1/4" sq. ( 4X4), so 16 X 200 is 3200#, (granted she is standing on 1 heel.) something to think about...
.... Right you are... I never did do very well in geometry XD Could throw algebra at me all day long, but for some reason, geometry just never stuck...
Except, gentlemen, psi isn't relative to the presses rating or to the way it increases power.
It is two different concepts and isn't usually used when talking about this type of work /tool.
If you are talking hydraulics, especially closed systems, then we are in business. And I can help a lot there. Thats what I get paid to do. Well, that and being eye candy.
Chilly