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Crash5291
08-11-2010, 01:02 AM
i am trying to figure the math out to determin how much force is needed for lift

this is a basic layout of my thought
"A" and "B" are on the fixed base and "C" is the arm

Now what the math that i use to figure out optimal "A" / "B" ratio (height offset) to obtain "X" lift pushing on "C"

i forgot to add "D" to the image its the lift point its 120" from "A" center line.
1000lb's is the target

Whats the thoughts guys?

Edit - My design is open to modification/rearrangement

socoj2
08-11-2010, 07:15 AM
F * Force Arm = Resistance * Resistance Arm

kenklingerman
08-11-2010, 10:22 AM
ok, you have 1000 lbs 120" (10 ft) from pivot "A", thats 10,000 ft/lbs
Now you need to create an apposing moment arm that exceeds 10,000 ft/lbs in the opposite direction
about pivot "A". to do this you need to lay in the force from pivot "B" to pivot "C". now you can create and measure the arm from pivot "A" to 90 degrees to force vector "BC". My layout indicates 26.35 inches (2.2 ft) while the system is in this position. So divide 10,000 ft/lbs by 2.2 ft to get the required force to match the moment in the opposite direction about pivot "A" which is 4545.45 lbs.

remember though that as the lift raises the moment arms change.

PS your layout does not appear to be to scale, I could be wrong though.

Crash5291
08-11-2010, 07:11 PM
No layout is not to scale just a representation of the initial design

Thanks for the help

so at 90 Degrees a 1 1/2" ram with 2575psi would net the required lift?

Thanks this helped me figure it out i think now to just fudge with the distances to get something i like

kenklingerman
08-11-2010, 11:38 PM
No layout is not to scale just a representation of the initial design

Thanks for the help

so at 90 Degrees a 1 1/2" ram with 2575psi would net the required lift?

Thanks this helped me figure it out i think now to just fudge with the distances to get something i like

barely but yeah, I would assume you would want some extra lift though. It does sound like u understand that part of the math anyway:-)

By the way, what I just did is called a free body diagram, not to hard to learn, look it up, if you enjoy physics, you'll like learning about it!

William McCormick Jr
08-11-2010, 11:50 PM
I would calculate that you need 9,000 pounds from your piston just to hold 1,000 pounds. Not lift or hold it safely, just balance it out there.

The point that you mount the piston to the crane, not the arm or boom, is the distance that you have to calculate too. So you are actually lifting from 12 inches out. Making the weight you are going to lift seem 9 times heavier then the 1,000 pound payload. According to my calculations.

Sincerely,

William McCormick

kenklingerman
08-12-2010, 12:43 AM
I would calculate that you need 9,000 pounds from your piston just to hold 1,000 pounds. Not lift or hold it safely, just balance it out there.

The point that you mount the piston to the crane, not the arm or boom, is the distance that you have to calculate too. So you are actually lifting from 12 inches out. Making the weight you are going to lift seem 9 times heavier then the 1,000 pound payload. According to my calculations.

Sincerely,

William McCormick

with all due respect, the moment from vector "BC" to pivot "A" is 26.35 inches, not 12, you have to calculate the moments at the correct point of application. Lay it out, I did very carefully..... and yes my numbers represent balancing it out, thus my second reply....

Crash5291
08-12-2010, 12:47 AM
i don't quite follow you William, what i read of your post seems like how it is worded above

i did a scale drawing and some measurements

Point A to C is 36.497" @ 350.5 Degrees angle
Point B to C is 30.610" @ 218.4 Degrees angle
Point A to B is 27.731" @ 295.6 Degrees angle
all are center to center on the pin.

William McCormick Jr
08-12-2010, 02:04 AM
Here is a video of the reason why you need more piston power. It is interesting after the halfway mark, you still need 2,000 pounds to hold the boom, all the way out to the end of the boom. That is a phenomena.

Sincerely,

William McCormick

socoj2
08-12-2010, 07:12 AM
Do not pay attention to william.

Its akin to feeding trolls.

kenklingerman
08-12-2010, 10:55 AM
OOPs just dug out my statics book and discovered I didn't do that quite right, the correct answer to balance the load of 1000 lbs is 5555 lbs/force along "BC" Sorry...

Crash5291
08-12-2010, 05:07 PM
whats the difference in the math?

kenklingerman
08-12-2010, 11:26 PM
whats the difference in the math?

I forgot to create a vector diagram, at "C" the force to balance is straight up is 3333 lbs, but the cylinder is pushing at an angle that has to be vectored in, I dug out my ol' college statics book and went OOPS and re-did it according to the text

it's all good now, I'm sure 6000 lbs along BC will be plenty!

William McCormick Jr
08-12-2010, 11:37 PM
OOPs just dug out my statics book and discovered I didn't do that quite right, the correct answer to balance the load of 1000 lbs is 5555 lbs/force along "BC" Sorry...

That seems like it is in the ball park to determine the straight up and down load on "C" when the boom is horizontal.

But I believe you also have to then calculate the lack of tangent pressure and movement against "C" pivoting on "A", from "B".

My guess would be that you are back at around 10,000 pounds. Just to hold it there with no margin of safety.

Cranes and booms are great fun to calculate. You are always going to make mistakes. I made two different movies, one I tossed out, after I caught a mistake in the first one.

There is also a lot of friction generated at pivot points, mounting points, and there is even a flexing of the material that has to be taken into consideration. It can cause what would seem like the right piston, to freeze up for a fraction of a second as it goes to transfer power to the boom. Making it seem like an undersized piston that it actually becomes.

Also consider that when you have a crane that it will often be used to glide an object into place. So that even though the boom is almost upright, there can be a tremendous angle created in the cable or chain from the boom to the object you are lifting. On purpose to position an object being lifted. It often takes very little pressure to do it. Or a big wind can come along and just do that by accident.

Normally I see it like, where the piston is mounted, to the crane, that should be about an accurate place to make a guess about how much weight you need to balance out the boom. Then calculate in everything else, and test it.

Sincerely,

William McCormick

William McCormick Jr
08-13-2010, 02:07 PM
I forgot to create a vector diagram, at "C" the force to balance is straight up is 3333 lbs, but the cylinder is pushing at an angle that has to be vectored in, I dug out my ol' college statics book and went OOPS and re-did it according to the text

it's all good now, I'm sure 6000 lbs along BC will be plenty!

If you look at the scales of justice, you realize that if you are weighing an object, on one side of the scales, in order to hold the two sides of the scales level, you will have to apply a similar force down upon the opposite side of the scale.

When you do that, you transfer twice the weight of the object you are weighing to the center point holding the object being weighed and the counterweight. That center point, now holds the weight of the object you are weighing, plus the counter balance.

So if we have 1,000 pounds out on a ten foot boom, we know that at five feet there is a 2,000 pound load. Since we know the load at five feet is 2,000 pounds, we know that at two and a half feet out on the boom holding 1,000 pounds, there is a force of 4,000 pounds. No two ways about it, nothing to think about. You can easily demonstrate that at home with simple objects.

So I know that at 30 inches, there is 4,000 pounds of pressure. At 24 inches there has to be more pressure then 4,000 pounds.

You might be able to see in the attachments that the vertical "H" beam obviously has fifty pounds added to its own weight. Even though the "I" beam has only 25 pounds five feet out on the "I" beam. The "H" beam represents the Point "C" on the crane. Point "C" becomes a pivot point, a leverage point. You have to calculate for that.

Sincerely,

William McCormick

William McCormick Jr
08-13-2010, 09:35 PM
I think you will like this a lot guys. I could not believe how much weight it took to balance the balsa wood boom in this movie. That was more fascinating to me then anything else. I would have bet money that it would not take that much weight to do it.

But you can see it is pretty much like I stated that as you half the distance from the payload to the crane, you double the downward force upon the boom at that point. And if you move the point again to half the distance to the crane, You will double the amount of downward pressure upon the boom at that point.

This is wild stuff.

Sincerely,

William McCormick

Crash5291
08-14-2010, 02:09 PM
HA!

Got it sorted now it all makes sense in my head and i am Smoking on it.

I Will update this when i get it all sorted out

steve45
08-14-2010, 05:27 PM
William, I really don't follow your logic, "Experiment, build multiple prototypes, and then test, and test some more... Anyone can be wrong about crane design."

This is why we have engineering.

William McCormick Jr
08-14-2010, 07:25 PM
William, I really don't follow your logic, "Experiment, build multiple prototypes, and then test, and test some more... Anyone can be wrong about crane design."

This is why we have engineering.

I am just highlighting things that I have seen supposed engineers screw up and screw up others with. I talk to engineers a lot, and they have some far out hieroglyphic like calculations that very often come to the wrong conclusions.

What I am demonstrating in that movie requires no real rocket science, it is a simple ratio. A ratio that anyone can understand and use. And you can become more comfortable with understanding it at home using simple items around the house.

Over the years I have had engineers tell me that a man holding in one hand a rope, through a pulley to a 25 pound weight being held off the ground, and in the other hand a rope through a pulley holding a 25 pound weight off the ground. Picture below. Puts the man under 25 pounds of force. Of course the fellow is under fifty pounds of force to tear his body apart.

Which means that all of the rope involved must also be experiencing an actual 50 pounds of force. Engineers say nay. They say we measure tension without taking any of that into consideration. What saves them is the actual rating on the rope, the rating on the rope is in pounds the rope can lift. That is a build of a prototype, an experiment or experiments, and test and retest. That was not just engineering. That test does not show what force the rope is under. That just shows a specific force from a specific object at a specific weight, and velocity, that the rope can handle. That is why today you can find some links and parts within lifting assemblies that are undersized.

Some of these same engineers claimed that a cantilever does not double the weight of the beam or floor joist, resting on the wall supporting the cantilever, as opposed to the beam or floor joist supporting a continuous wall going straight up with no cantilever. Both walls being of equal weight, one cantilevered.

So you can show me college degrees, all you like. Until I hear some hard understanding of these very strange effects, I will not be quick to take college engineering seriously.

Look at the stuff being made today.

The pictures below are not trick pictures, you can argue the wording till the devils hot tub freezes over. Because depending on, from what angle you are talking about it, any of the below pictures, can have more then one meaning. But I have been arguing these points for years and some engineers really do get it. And they get a kick out of the fact that they never thought about it like that before. Yet older hands on engineers taught me this stuff.

This goes back to Newtons equal and opposite forces necessary to hold an object under the influence of gravity still.

Twenty five pounds free falling is 25 five pounds of force. To stop it you need an additional 25 pounds of force in the opposite direction. Creating 50 pounds of force to hold it still. Neither force stops while a 25 pound object sits still. Both forces are totally active and effective. That is universal science.

Sincerely,

William McCormick

SolidWelder
08-23-2010, 10:41 PM
William:
I have read many of your posts over the last year or so and realize that you are a veritable font of practical knowledge when it comes to many everyday construction issues. When it comes to engineering statics, however, I believe that you really should seek to understand these concepts better than you seem to, before providing such strong arguments as to why things are so, and why the engineers are all wrong and you are right.

There is only a 25 pound tension force in a persons body, when pulled with 25 lbs from both sides. If you want to understand why, google "Free Body Diagram", and focus on the concepts of an external force acting on a body, versus a body's internal forces. I'm sure that at some point you will get it, and understand why it is so.

Also, consider a man, with his arms horizontal being pulled on his left arm with 25 lbs and his right arm grabbing and holding tight to a fixed grab bar. What force is in the man's torso then. Answer : 25 lbs tension.
Consider this: What is the difference between (a) his right arm pulling on the grab bar with 25 lbs force vs. (b) the grab bar "pulling him" with 25 lbs force. Ans: None- Each are exerting an equal and opposite force of 25 lbs on each other.
Hope this helps you reconsider your opinions on this.

ZTFab
08-23-2010, 11:33 PM
William:
I have read many of your posts over the last year or so and realize that you are a veritable font of practical knowledge when it comes to many everyday construction issues. When it comes to engineering statics, however, I believe that you really should seek to understand these concepts better than you seem to, before providing such strong arguments as to why things are so, and why the engineers are all wrong and you are right.

There is only a 25 pound tension force in a persons body, when pulled with 25 lbs from both sides. If you want to understand why, google "Free Body Diagram", and focus on the concepts of an external force acting on a body, versus a body's internal forces. I'm sure that at some point you will get it, and understand why it is so.

Also, consider a man, with his arms horizontal being pulled on his left arm with 25 lbs and his right arm grabbing and holding tight to a fixed grab bar. What force is in the man's torso then. Answer : 25 lbs tension.
Consider this: What is the difference between (a) his right arm pulling on the grab bar with 25 lbs force vs. (b) the grab bar "pulling him" with 25 lbs force. Ans: None- Each are exerting an equal and opposite force of 25 lbs on each other.
Hope this helps you reconsider your opinions on this.

You may want to rethink that part of it.

William McCormick Jr
08-24-2010, 02:40 AM
William:
I have read many of your posts over the last year or so and realize that you are a veritable font of practical knowledge when it comes to many everyday construction issues. When it comes to engineering statics, however, I believe that you really should seek to understand these concepts better than you seem to, before providing such strong arguments as to why things are so, and why the engineers are all wrong and you are right.

There is only a 25 pound tension force in a persons body, when pulled with 25 lbs from both sides. If you want to understand why, google "Free Body Diagram", and focus on the concepts of an external force acting on a body, versus a body's internal forces. I'm sure that at some point you will get it, and understand why it is so.

Also, consider a man, with his arms horizontal being pulled on his left arm with 25 lbs and his right arm grabbing and holding tight to a fixed grab bar. What force is in the man's torso then. Answer : 25 lbs tension.
Consider this: What is the difference between (a) his right arm pulling on the grab bar with 25 lbs force vs. (b) the grab bar "pulling him" with 25 lbs force. Ans: None- Each are exerting an equal and opposite force of 25 lbs on each other.
Hope this helps you reconsider your opinions on this.

If you look at the one foot radius wheel that has two 25 pound weights attached to it. You will probably agree that they create 50 foot pounds of force.

So we can conclusively throw away the idea that two 25 pound weights create only 25 pounds of force.

This will show up in any joint under pressure, that is not perfectly geometric and secure. If you have a connection rated at a certain load. And the attachment can drift left and right, at the connection points, that allows a non-symmetrical force to be created and many common links do allow this movement. You could find that you have placed upon the housing, a force that is double the normal force expected. And perhaps 10 times less then what was calculated for.

A good crane operator will often center a lifting cable at a connection. Other guys will laugh and not understand what the big deal is. He has been properly instructed that the link is designed to be used from its center. He may or may not understand the why.

If we hang two fifty pound weights from the ceiling we know the ceiling is lifting 50 pounds of weight. Again we can agree upon that.

If a guy is holding two 25 pound weights off the ground we know his body is under fifty pounds of force. Or you are saying he can lift 50 pounds with 25 pounds of force.

If you look at the scrap scale, I know I want fifty pounds worth of credit, for the metal in those two 25 pound pails of scrap metal. Not the 25 pounds that the scale reads.

It is funny but in my area everyone used to be like me. I was taught by people that laughed at colleges, and conclusively exposed colleges as somewhat fraudulent. My friends even inspired good colleges to work up statistics, of the average eighth grader with a few years of apprenticeship under his belt, compared to the fresh collegiate mind master, and an imagination of how things really work.

The honest colleges reported horrific results against colleges success. They estimated at the present rate that many college students would be needed to work harder, to produce less then one well trained apprentice with an eighth grade education. These colleges immediately lost government grants. Some claimed it was because their statistics were wrong.

So it is where you were raised. What you were taught and what you would like to believe. Colleges agreed with government that everyone had to have the same information even if it was wrong. And we knew it was wrong. But there was big money paying for wrong to be right. Stupid money. Ha-ha.

Here is a common thing that you can actually witness now and then. The spool coming out of a winch. Some winch companies actually test their stuff. And they know what I am saying, but why open their mouths. If they do they are calling for rebellion against anarchy. Better they just keep what works to themselves and stay low. Picture below.

The funny thing is, that there was always two hundred pounds there, the lever or sea saw, just highlights it.

I am not telling you to throw away everything you ever learned. But rather just properly label what you learned with the right numbers.

There are good old timers that will tell you pretty much what I am saying is correct. Because they had the broken pieces of metal in their hands, the metal that could not have broken under that load. These guys are under pressure, and they are still willing to take the time and explain it to you if you wish to endeavour to those extremes.

Sincerely,

William McCormick

steve45
08-26-2010, 08:40 AM
William:
I have read many of your posts over the last year or so and realize that you are a veritable font of practical knowledge when it comes to many everyday construction issues. When it comes to engineering statics, however, I believe that you really should seek to understand these concepts better than you seem to, before providing such strong arguments as to why things are so, and why the engineers are all wrong and you are right.

There is only a 25 pound tension force in a persons body, when pulled with 25 lbs from both sides. If you want to understand why, google "Free Body Diagram", and focus on the concepts of an external force acting on a body, versus a body's internal forces. I'm sure that at some point you will get it, and understand why it is so.

Also, consider a man, with his arms horizontal being pulled on his left arm with 25 lbs and his right arm grabbing and holding tight to a fixed grab bar. What force is in the man's torso then. Answer : 25 lbs tension.
Consider this: What is the difference between (a) his right arm pulling on the grab bar with 25 lbs force vs. (b) the grab bar "pulling him" with 25 lbs force. Ans: None- Each are exerting an equal and opposite force of 25 lbs on each other.
Hope this helps you reconsider your opinions on this.
Agreed. I'm planning a reply to this, but haven't had time to do the graphics to explain it.

William McCormick Jr
08-30-2010, 12:02 AM
Agreed. I'm planning a reply to this, but haven't had time to do the graphics to explain it.

Before you do that check out these two pictures well. You can see conclusively that we have the same two twenty five pound weights in both cases. So we can put this aside, we know, or at least we agree that in both pictures there are two separate twenty five pound weights hanging on two separate ropes.

When we look at the one foot radius wheel, we immediately know that there is fifty foot pounds of force being placed upon the wheel from the two twenty five pound weights.

The reason it is obvious, and we can agree there is fifty pounds of force, is because we know that two twenty five pound weights create 50 pounds of force at one foot diameter.

The wheel and all the fancy leverage does not actually change anything. It just shows or gives us a way to highlight the force that was always there. There is always 50 pounds of force created by two twenty five pound weights, under the influence of gravity.

But schools have apparently taught otherwise. So much so that it takes a person put through school some time to go back and reorganize his thoughts about this. He has been told and shown math that says the fellow holding two twenty five pound weights only experiences twenty five pounds of force. However if you study the one foot radius wheel with two twenty five pound weights you see that there is certainly 50 pounds of force created by two twenty five pound weights. No tricks no gimmicks just hard reality.

Every time I come back to this, I have to go through my notes and make sure I am correctly explaining it. It is a very tricky subject. And our language about such things lends itself to error.

Sincerely,

William McCormick

kenklingerman
08-30-2010, 12:54 AM
Sorry but how does this relate to the OP's original question? WTF is up with the blow up sex doll any way? IT is just a statics question, engineers may be dumb asses when it comes to common sense sometimes but trust me they do know their statics, if nothing elsem they wouldn't have their degrees if they didn't.

William McCormick Jr
08-30-2010, 08:43 PM
Sorry but how does this relate to the OP's original question? WTF is up with the blow up sex doll any way? IT is just a statics question, engineers may be dumb asses when it comes to common sense sometimes but trust me they do know their statics, if nothing elsem they wouldn't have their degrees if they didn't.

That little doll is Mr. Bill, part of a comedy skit on Saturday Night Live. He always got smashed at the end of the skit.

It is not as cut and dry as you might believe. If you saw the balsa wood beam in that movie I posted, and the tremendous weight it took to counterbalance a very light piece of balsa wood slightly off center. You probably will realize that a heavy boom alone creates massive forces upon the crane holding the boom. And even more force upon the point moving the boom up and down. This has everything to do with the original question.

If you cannot easily understand the doubling effect of the lever, and how it highlights the actual and total force a load places upon a crane. You cannot build a crane safely.

Many engineers really do not get it. They talk a good story, however they do not get it. I have seen things blow apart. Usually it is the object, that I am highlighting, that is actually under a double load.

You will note my estimates are the actuality for just holding the boom still. Not really lifting it under the load quoted. I have reality on my side.

A scale is calibrated to measure exactly half the force it is under. I am trying to highlight that as you build things for a given load. That actual forces, that we do not normally think about in our toy building block kind of life. Suddenly are sprung upon us in the real world. I watch stuff fail all the time. I watch experts bend equipment. All because they did not know the actual weight they were applying to an object.

The engineers calculations often lack what I am demonstrating here. Most engineers do not feel confident with what I am talking about. That is why most do not post and offer their opinion to the problem. Or shoot me down.

I work with engineers and I am sure they are lacking confidence in this area. I know they have had failures. And what they do is they learn from the failures. And now they have a whole bunch of 2X or 3X or 4X rules. Added on to the original math that failed them. That is crazy. That is the end of our civilization.

I would be loud as heck about it. And condemn the "majority engineering think" about this subject. Kind of like I am doing now.

This should be a fun subject that any man that wants to build just about anything should understand. And he should be able to pass this along to his kids.

The one foot radius wheel tells it all. If we apply two 25 pound weights to the one foot radius wheel, we can hold 50 pounds. So there is fifty pounds of force created by two 25 pound weights. The lever brings this out in an often comical way. A very powerful way. An adjustable boom is a lever. It is also a wheel for engineering purposes, and foot pounds apply. This is one and the same. You cannot understand one without the other.

I would not be quick to praise college static formulas. Look around.

I have seen college engineers claim fantastic fantasy, is better to explain the failure of a project, rather then admit their math and formulas are not taking into account the reality at hand.

I could make a simple formula to calculate the force at any given point on a crane boom. Using nothing but a few simple ratios. Based on my test booms. It would not be rocket science. But rather tough rugged American Garage Mechanic style thinking.

Many people do not know what they are doing or building.

Sincerely,

William McCormick

steve45
08-30-2010, 11:34 PM
Two different problems, William. You're equating a moment to a straight line force.

If you were to attach the two 25 pound weights to the same point on the wheel, the forces would cancel out without the 50 pound weight.

Donald Branscom
08-31-2010, 11:41 PM
Here is a video of the reason why you need more piston power. It is interesting after the halfway mark, you still need 2,000 pounds to hold the boom, all the way out to the end of the boom. That is a phenomena.

Sincerely,

William McCormick

For every foot out from the pivot or beam securing point, load is multiplied times 1,000.
Do not forget to add in a safety factor.
Static as opposed to a dynamic load.

tkanzler
09-01-2010, 01:59 PM
ok, you have 1000 lbs 120" (10 ft) from pivot "A", thats 10,000 ft/lbs
Now you need to create an apposing moment arm that exceeds 10,000 ft/lbs in the opposite direction
about pivot "A". to do this you need to lay in the force from pivot "B" to pivot "C". now you can create and measure the arm from pivot "A" to 90 degrees to force vector "BC". My layout indicates 26.35 inches (2.2 ft) while the system is in this position. So divide 10,000 ft/lbs by 2.2 ft to get the required force to match the moment in the opposite direction about pivot "A" which is 4545.45 lbs.

remember though that as the lift raises the moment arms change.You were right the first time, and you're also right that the geometry changes with angle - the cylinder force increases a bit off the horizontal, though not hugely, then starts to drop off. A graphical solution is a perfectly valid way to analyze this, along with several vector math methods.

OP: None of this takes into account the dead load of the beam, proper beam sizing and connection design (including weld joint size, type, and design), appropriate design criteria for hydraulics, loading and stability due to side loading, impact loading, load drift, reactions at the various pins and resulting design of interfaces, overall stability of the crane (if stability based) or reaction at the supporting structure (if strength based), material selecton, the list goes on and on.

William McCormick Jr
09-01-2010, 09:55 PM
Two different problems, William. You're equating a moment to a straight line force.

If you were to attach the two 25 pound weights to the same point on the wheel, the forces would cancel out without the 50 pound weight.

If you attach both 25 pound weights to the same point on the wheel, then you just take the wheel out of the picture, or out of the equation.

I am putting the force of two twenty five pound weights through the wheel to show how much force they are actually delivering to the wheel. So we can actually measure the force of two twenty five pound weights.

The force from the two twenty five pound weights does not change. Two twenty five pound weights create fifty pounds of force. It is just lucky we measure rope by a design build test method. And that we did not leave it up to just paper engineering. Or we might be in even worse shape.

Your explanation highlights a majority of American citizens that have never been taught or shown, if you can create 50 pounds of torque with two twenty five pound weights attached to a point 1 foot away from the pivot point of the wheel. Then you can be sure that two 25 pound weights can create 50 pounds of torque at one foot radius. Which is the same force as a straight line force. A gear bar converts foot pounds of torque to straight line force at one foot radius.

If you have 50 foot pounds of torque from a one foot radius gear, the gear bar attached to the gear, will give you 50 pounds of straight line force. If you hang two twenty five pound weights on a vertical gear bar, a one foot radius gear attached to the gear bar will have 50 foot pounds of torque available to turn or move something else.

I am not using the wheel for leverage, rather as a way to see the actual force on the rope from two twenty five pound weights. The same weight or tension was always on the rope, you just did not know it. The wheel with two twenty five pound weights pulling the wheel in the same direction, at the same radius as the fifty pound weight, creates no unbalanced ratio of movement. The movement is one to one. You are just looking at the force of two 25 pound weights.

There is no gearing involved. I am just using that example to highlight what force the rope is actually under.

In that wheel diagram in the previous post, you have the two 25 pound weights pulling from a point tangent to the wheel. So it becomes a straight line force, much like a gear bar. Really check it out for a while, it kind of just suddenly falls into place.

Think about two hanging scales, hanging from a ceiling. Each scale has twenty five pounds in it. We know for sure, no doubts, the ceiling is under a fifty pound load. We can easily agree on that. It takes two ceiling scales, to correctly measure both 25 pound loads. So we add the two scales together and come up with the total weight the ceiling is supporting.

Now if you put two scales between two twenty five pound weights each will read 25 pounds, totaling 50 pounds of force on the rope. We saw with the scrap yard, comedy scale, that you cannot properly measure the two twenty five pound weights with one scale. So we need two scales to actually see what force the rope is under from two 25 pound weights. That is just how you have to design scales. They are designed to read only one half the force they are actually under.

Believe me many individuals cannot even discuss this subject, the fact that you can is almost remarkable.

Sincerely,

William McCormick

steve45
09-02-2010, 06:01 PM
William, you could put a hundred scales in place of the two shown in your drawing (end-to-end, of course) and they would all read 25 pounds. Does that mean you have 2500 pounds of force applied?

William McCormick Jr
09-02-2010, 09:41 PM
William, you could put a hundred scales in place of the two shown in your drawing (end-to-end, of course) and they would all read 25 pounds. Does that mean you have 2500 pounds of force applied?

If you really wish to understand this. Look at the scrap yard scale. We know for sure that one scale does not properly measure two suspended 25 pound weights.

Now look at the hanging scales suspended from the ceiling, we need two scales to measure, the two twenty five pound weights. And we have to add the two scales together to get the weight on the ceiling.

So if you have two twenty five pound weights with a single scale between them, we are sure the scale does not properly measure, the two twenty five pound loads. If we have two scales suspended from the ceiling, we can be sure, there is fifty pounds on the ceiling and to come to that conclusion we have to add the two scales together.

The scale only measure half the force upon it. Just look at the comedy junk yard scale.

Lastly look at the third picture, and you realize that is the same as the two scales hanging from the ceiling. So we must add the two scales together to know how much force is upon the rope.

Sincerely,

William McCormick

steve45
09-02-2010, 10:53 PM
If you really wish to understand this. Look at the scrap yard scale. We know for sure that one scale does not properly measure two suspended 25 pound weights.

Now look at the hanging scales suspended from the ceiling, we need two scales to measure, the two twenty five pound weights. And we have to add the two scales together to get the weight on the ceiling.

So if you have two twenty five pound weights with a single scale between them, we are sure the scale does not properly measure, the two twenty five pound loads. If we have two scales suspended from the ceiling, we can be sure, there is fifty pounds on the ceiling and to come to that conclusion we have to add the two scales together.

The scale only measure half the force upon it. Just look at the comedy junk yard scale.

Lastly look at the third picture, and you realize that is the same as the two scales hanging from the ceiling. So we must add the two scales together to know how much force is upon the rope.

Sincerely,

William McCormick

I DO understand this, William. Sorry, you are wrong. If you attach one end of your 'junk yard scale' to a wall, it will read 25 pounds, which is the correct reading. If you replace the wall with another 25 pound weight, it will read 25 pounds, which is the correct reading. Why would the scale lie? It doesn't know what it's connected to. It only reads the load that it sees. Think of it like this, the scale isn't reading the total weight, it is reading the tension in the line. There is a difference.

So answer my question: If you replace the single scale, or the double scales, with a bunch of scales, what do you think they would read?

Hint: the scales would all read 25 pounds!!!

Boostinjdm
09-03-2010, 12:13 AM
I DO understand this, William. Sorry, you are wrong. If you attach one end of your 'junk yard scale' to a wall, it will read 25 pounds, which is the correct reading. If you replace the wall with another 25 pound weight, it will read 25 pounds, which is the correct reading. Why would the scale lie? It doesn't know what it's connected to. It only reads the load that it sees. Think of it like this, the scale isn't reading the total weight, it is reading the tension in the line. There is a difference.

Thank you! I've been trying to figure out how to word a response since he first posted that example.:drinkup:

William McCormick Jr
09-03-2010, 09:10 PM
I DO understand this, William. Sorry, you are wrong. If you attach one end of your 'junk yard scale' to a wall, it will read 25 pounds, which is the correct reading. If you replace the wall with another 25 pound weight, it will read 25 pounds, which is the correct reading. Why would the scale lie? It doesn't know what it's connected to. It only reads the load that it sees. Think of it like this, the scale isn't reading the total weight, it is reading the tension in the line. There is a difference.

So answer my question: If you replace the single scale, or the double scales, with a bunch of scales, what do you think they would read?

Hint: the scales would all read 25 pounds!!!

Of course fifty scales all chained together will all read 25 pounds if the last one is holding the twenty five pound weight. Give or take for the weight of the scales.

However your claim that you understand this is not yet true. Then actual force upon the rope holding a scale. Whether the rope is tied to another twenty five pound weight or a wall. Has fifty pounds of force upon it. That is just how it is.
A wall exerts a counter force of 25 pounds, against a twenty five pound weight to hold it from falling. So you have fifty pounds of actual force upon the rope holding a twenty five pound weight.

I only use wheels and levers, to put the force through those mechanisms, to show you the hidden force that one twenty five pound weight actually creates when held still. It was always there most do not see it or calculate for it.

A scale just shows you the weight of the object being weighed. It does not at all tell you the force upon the scale or rope holding the scale. If you want to know that just double what the scale says.

If you saw the picture of the winch, that has a one hundred pound load, upon it. Most would believe that is all the force there is. However after careful examination you realize that one hundred pounds actually created 200 pounds of force in the winch. Because of the equal and opposite force necessary to hold a 100 pound force still.

So the scale only measures half the force present from a weight placed on the scale.

In the case of the comedy junk yard scale, it does not measure both loads upon it. It measures one load only. Because the scale is designed to indicate only half the force it is actually under.

The comedy junk yard scale is holding two twenty five pound weights off the ground that requires 50 pounds of force in my universe.

And the two ceiling mounted scales each with a twenty five pound load in them concur with that way of thinking.

Look at the scales of justice, a balancing scale. You put an equal number of known or trusted, standard, ounce or pound weights, basically on one side of a sea saw, until it is level, to measure the weight of the load you want to measure. Now if we put a twenty five pound load to be measured, on one side of the scale, and then we put twenty five pounds of counterweight on the other. I know we have fifty pounds of force, pressing down on the scale. Now the whole scale weighs 50 pounds more then the scale itself.

So we know that in the comedy junk yard scale scenario, that the scale is actually under 50 pounds of force, however it is designed to indicate half that amount. If you hold a twenty five pound weight still with a rope. The rope has actually 50 pounds of force upon it. We just luckily measure rope with a design, build, test, logic. That just tells us that a rope designed for a twenty five pound weight, will hold a twenty five pound weight still. It does not tell us how much force the rope is under.

When you weld a cross member or other stiffener in place you may want to keep in mind what you are actually doing and the force you may be creating with your structure. It may very well be twice what you believe it is.

Sincerely,

William McCormick

steve45
09-04-2010, 12:13 AM
OK, William, this will strain your brain. Two pulleys, with a scale between them, with weights hanging down on each side (as shown in your example). HOWEVER, this time have a 25 pound weight on one side, a 50 pound weight on the other side, and let it rip! The cables are considerably long, what will the scale read???

Boostinjdm
09-04-2010, 12:27 AM
Dammit steve! Don't encourage him.

25 lbs.....

William McCormick Jr
09-04-2010, 03:12 PM
OK, William, this will strain your brain. Two pulleys, with a scale between them, with weights hanging down on each side (as shown in your example). HOWEVER, this time have a 25 pound weight on one side, a 50 pound weight on the other side, and let it rip! The cables are considerably long, what will the scale read???

It could read nearly up to fifty pounds until the 25 pound weight accelerates, to the speed of a free falling object. Then the scale will read 25 pounds.

I have done electronically timed experiments of different weight objects, free falling objects, connected to one another over a pulley. To see the rate at which they accelerate. I was surprised at the velocity created, by two slightly different weights being hung over a pulley with a very good bearing.

Engineers miss weight verses stress. This is because laymen in the field really shook them up, when the laymen were using the, design, build, test, method of rope, cable or chain, to determine the size of rope, cable or chain they needed. It took out a lot of the, paper engineering work. However not in all cases was it effective. In some cases it only supplied half the cable, chain or rope strength needed.

But since the average dock worker could just do the calculations in his head for rather complex projects. The engineers decided to package that way of thinking. Unfortunately they did not even understand their own science of stress and force, so they did not add in the proper rules for when and where the simple formula would, and would not work. Or how it worked.

Sincerely,

William McCormick

tkanzler
09-04-2010, 03:37 PM
:cry:Engineers miss weight verses stress. This is because laymen in the field really shook them up, when the laymen were using the, design, build, test, method of rope, cable or chain, to determine the size of rope, cable or chain they needed. It took out a lot of the, paper engineering work. However not in all cases was it effective. In some cases it only supplied half the cable, chain or rope strength needed.

But since the average dock worker could just do the calculations in his head for rather complex projects. The engineers decided to package that way of thinking. Unfortunately they did not even understand their own science of stress and force, so they did not add in the proper rules for when and where the simple formula would, and would not work. Or how it worked.Dude!

Yikes!

I mean, like . . . yikes!

I'm speechless.

I guess I should go burn my engineering degrees, my P.E. registrations, and condemn every high-profile job I've ever done before someone gets hurt because my wire rope loads, structural members, and hydraulic pressures are really twice what I think they are.

And who knew that all those other engineers I interface with ever day were just humoring me all along. Or are similarly incompetent. Bummer. :(

Really!

Yikes!

:cry:

steve45
09-04-2010, 10:45 PM
I guess I should go burn my engineering degree...

I'm keeping mine, even though it's apparently worthless. Spent too much time & money to get it.

William McCormick Jr
09-06-2010, 02:17 PM
I have worked with really schooled engineers and they, definitely over complicate this subject. They overlook things totally obvious to a Janitor or laborer. Because the laborer has the reality on it.

I always ask the laborers if they think some temporary device we are building will hold or not. Not because I doubt what I did. But rather they have seen and lived through more accidents. They may recognize a potential danger that they have seen fail, one that I missed. Sometimes they will just tell you that you won't be able to turn a big object in the space that is there. This alone can save days of useless work. All because they have been there done that.

But the engineers that look like engineers, and some of the really scrubby unwashed ones, have stuff so beaten into their mind. That there is no way to calmly talk about it. There delicate memory of how to do it, is not capable of an interesting conversation about it. You can see that they have this delicate mind-set that is not subject to question. God help you, if you do question it. Ha-ha.

If I am leading a job and they have their mind set on something I know is wrong. I tell my guys that when it falls, to just watch it go, do not try to save it. Just stand their and laugh.

Sometimes I catch hell, but after a few times they realize that I knew what was going to happen. And the engineers were just experimenting, with their money. And my men's lives.

A really good engineer will only offer up truly hard won lessons that he is sure of. Often from that he can tell who is right and who is wrong, by seeing who is trying to break or avoid that rule he is sure of.

I know some engineers that know exactly what I know and they use simple language to express it. They know there is 50 pounds of force on a rope holding a 25 pound weight still. No doubts about it. If you look back over the posts, you can see that many are fighting the simplicity. I go along with the Navy on KISS, Keep It Simple Stupid. This whole subject is proof that some are not following this procedure.

That one picture of the two twenty five pound weights attached to the wheel one foot out from the center, holding a 50 pound weight still, tells the story of the actual force created by one 25 pound weight being held still by a solid object, or a 25 pound weight being held still by a 25 pound counter balance. That is a one to one scenario, no gearing or ration of movement involved.

This manifests itself in engineering all the time. It is a simple thing. That is not simply addressed.

Sincerely,

William McCormick

pandinus
09-07-2010, 02:47 AM
Don't worry about it.... some people used to think the world was flat, AND it was the center of the universe... (so did they think the universe was flat too???) turns out they were just making assumptions about things they didn't properly understand... these days MOST people are better educated about such things... but I'm sure you could still convince a few people out there that its all flat :confused: :laugh:

tanglediver
09-07-2010, 03:30 AM
It depends on how many dimensions you think in.
Hypothetically....of course. :rolleyes:
http://blogs.discovermagazine.com/cosmicvariance/2005/12/07/how-many-dimensions-are-there/

Pasto76
09-07-2010, 02:41 PM
"They know there is 50 pounds of force on a rope holding a 25 pound weight still"

this is a bizarre statement my man. Ive come up through the trades, so I understand your perspective, and some of it is getting lost in your messages. Ive also a science degree, the same science that puts people in space, diagnosis and cures diseases, runs power plants - its all physics and calculus. Calling that notations "hieroglyphics" is absurd. Something we learned in the army - you believing something is not a requirement for it to be true.

your above vid with the balsa wood boom is a good visual demo of the applied physics, but it doesnt replace the math involved.

be safe out there

timrb
09-07-2010, 11:54 PM
You guys *do* know that you're arguing with the most masterful troll ever to walk the Earth, right? We are, all of us here, in the presence of greatness.

Tim

ZTFab
09-08-2010, 12:06 AM
You guys *do* know that you're arguing with the most masterful troll ever to walk the Earth, right? We are, all of us here, in the presence of greatness.

Tim

Some of us know exactly who we're dealing with.

...and he's about one post away from never coming back.

joedirt1966
09-08-2010, 04:50 PM
I know some engineers that know exactly what I know and they use simple language to express it.

Sincerely,

William McCormick

Some people would go to town with the above quote.:laugh:

steve45
09-08-2010, 11:58 PM
William, I never refuse to listen to an experienced person's opinion just because they don't have a degree. I know a lot of people that have a lot of knowledge without formal training.

On the other hand, if someone tells me I'm wrong, I expect them to be able to explain why. For instance, when you to tell me that a scale is calibrated to read only half of the force that it sees, that is absurd. I don't care if somebody else agrees with you, either.

I worked in an engineering design department for a few months. The company 'jackhole' came by and criticized the exhaust system I designed for a 4500 HP engine, saying that it was far too large, expensive, and it didn't need to be designed to allow movement with expansion. He had no formal training in engineering, and when I showed him the manufacturer's specs, the flow calculations, thermal expansion calculations, etc., he just declared it BS and went over my head to get my design dropped. After I left the company, one of the other engineers told me that the revised design was a miserable failure. Components broke apart and the engine didn't perform as it should. After considerable embarrassment and expense, they dug my original design out of the files and used it--successfully...

As one of my engineering professors used to say, "If you can't explain a design problem mathematically, you don't know what you're talking about."

The defense rests.

William McCormick Jr
09-09-2010, 12:35 AM
Some people would go to town with the above quote.:laugh:

Depending on where you are looking at that from, that could be funny.

A practical example that I have seen here on the forum, that over looks some of these principles we have discussed, in this topic. Is allowing welders to run partial welds on gussets. Now if I had asked them if they measure exactly how far, from the edge of the gusset do they start the weld, I would have been laughed at. Because that is just ridiculous, or is it?

However if a person was to start a weld just a half inch further in from the end of a gusset, that has a twin gusset on the other side of what it is supporting. The other twin gusset could double the expected load upon the weld of the gusset with the weld started in further from the end of the gusset.

Just trying to let you guys in on the real deal. So you are not the fool or scapegoat of bad engineering. Do you think the engineer that overlooked this, isn't going to blame your quality weld, when it fails at half the strength the engineer expected it to hold. Ha-ha.

We have these arguments in the real world I know how they go. The engineers will ask you if that half inch matters. You tell them heck yes.

Sincerely,

William McCormick

kenklingerman
09-09-2010, 08:50 PM
Omg! Wtf!

ZTFab
09-09-2010, 08:57 PM
I've seen, heard and read enough from Billy Mac.

You are gone.

Your constant ramblings, mis-information, and attempts to change the subject when asked to prove your "theories" are over!

Go blow smoke up someone else's a\$\$!

pinjas
09-09-2010, 10:10 PM
I've seen, heard and read enough from Billy Mac.

You are gone.

Your constant ramblings, mis-information, and attempts to change the subject when asked to prove your "theories" are over!

Go blow smoke up someone else's a\$\$!

You seriously banned William?

ZTFab
09-09-2010, 10:50 PM
You seriously banned William?