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Re: determined metal thickness needed

Originally Posted by Ingenuity

I have been played by a fool.

I did some 'research' on the infamous 'William McCormick' and he/she/it is notorious on this site (and several others, namely, Miller, Hobart etc) on the insanity of this very subject (and several others).

This subject - along with the very same graphics that McCormick uses above - were the cause of his ban back in 2010: https://weldingweb.com/vbb/threads/4...igure-this-out

You can believe what ever you wish, but rejecting/disputing scientific fact is denialism.

Surely, after more than a decade of your 'crusade' on this subject, you must stop and think "maybe I am wrong".

Because you are wrong, William McCormick. Sincerely, wrong.

Then they realized I was right and here I am. When I first told people about helium allowing you to TIG weld much thicker aluminum with the same current starting around 2000 on the internet, I was an idiot. When I told people about welding aluminum, magnesium, and bronze with TIG straight polarity DC current to create a weld as strong as the material, I was a fool. Now it is just a common everyday thing. Soon people will know what I am saying about a cable holding a 25-pound weight off the ground being under 50 pounds of force as the reality. Some boob at a University will write a paper and take the credit for something that Archimedes knew.

I was talking to a very intelligent nurse about my back injury, and she highlighted how when I landed on my buttocks from a fall off a roof that my L1 and L2 vertebrae cracked and not my coccyx or lower lumbar vertebra. It is because of two forces the ground's upward force and gravities downward force. When my buttocks landed on the ground the only weight hitting the ground was my buttock's lower lumbar vertebrae. The weight from my head, arms, and shoulders couldn't immediately transmit to the ground as my spinal column absorbed the energy by compressing. The damage ended up at the spot that the two forces met, my L1 and L2 vertebrae.

Sincerely,

William McCormick

Re: determined metal thickness needed

Originally Posted by Ingenuity

I have been played by a fool.

I did some 'research' on the infamous 'William McCormick' and he/she/it is notorious on this site (and several others, namely, Miller, Hobart etc) on the insanity of this very subject (and several others).

This subject - along with the very same graphics that McCormick uses above - were the cause of his ban back in 2010: https://weldingweb.com/vbb/threads/4...igure-this-out

You can believe what ever you wish, but rejecting/disputing scientific fact is denialism.

Surely, after more than a decade of your 'crusade' on this subject, you must stop and think "maybe I am wrong".

Because you are wrong, William McCormick. Sincerely, wrong.

Evert time I pick up the subject I think I might be wrong. So I go through it from the beginning. And each time I do I come to the same conclusion. You cannot just explain away the counterforce necessary to weigh an object. If the two weights hanging from the ceiling scale are correctly recording 25 pounds and the ceiling has a load of 50 pounds upon it then surely the cable holding one or two weights has 50 pounds on it, due to the counterforce that is the same in both cases. The reason that cranes fail so often and cables are so large and still fail is that when a 10,000 pound load is jostled the stress on the rope goes from zero to 20,000 pounds not zero to 10,000 pounds.

Some people claimed I hurt their minds when I highlighted the reality, that they cannot explain away with logic, only math based on misunderstanding. Since I am a minority I tend to be voted away. Crane operators used to think as I do. Today I don’t know that all do but the good ones still do.

Sincerely,

William McCormick

Re: determined metal thickness needed

I have been played by a fool.

I did some 'research' on the infamous 'William McCormick' and he/she/it is notorious on this site (and several others, namely, Miller, Hobart etc) on the insanity of this very subject (and several others).

This subject - along with the very same graphics that McCormick uses above - were the cause of his ban back in 2010: https://weldingweb.com/vbb/threads/4...igure-this-out

You can believe what ever you wish, but rejecting/disputing scientific fact is denialism.

Surely, after more than a decade of your 'crusade' on this subject, you must stop and think "maybe I am wrong".

Because you are wrong, William McCormick. Sincerely, wrong.

Re: determined metal thickness needed

Here is another way to look at it. Now if I am correct, then the scales record half of the total force they are under when supporting an object. I am not suggesting chaos and madness, I am suggesting that most do not understand how a scale is actually calibrated. In the graphic below each hanging scale is correctly reporting the weight of the object hanging underneath it. Yet you can see the dilemma if the ceiling has a force of fifty pounds of downward force upon it. The hanging scale only reports the gravities effect on the weight, not the counterforce holding the scale-up.

Attachment 1730145

Sincerely,

William McCormick

Re: determined metal thickness needed

Originally Posted by Ingenuity

No, the upper cable has 180,000 lb, and each of the two lower cables has 90,000 lb - so combined, the two lower cables total 180,000lb. The sum of the vertical forces at the connection between the 2 lower cables to the 1 upper cable must = zero.





No, the scale does NOT measure half the force, it is measuring the only force that is applied to the system - 25 lb - that is in static equilibrium.


==> So you are not believing my math, and my graphics was not intuitive to you, so lets try this:


EXPERIMENTAL PROOF: The foundation upon which scientific theory is proven.



I used my DILLON tension gauge and a setup that mimics your 'scrap yard scale' pictorial:

First, measured the mass of the 2 x 25 lb weights to verify that they are indeed 50 lb total - measured in the vertical direction, using the DILLON gauge.

Photographic proof: A little hard to read in this photo, but it states 50 lbf. But I have other photos I can share if you need further proof. And the weights are stamped with 25lb - and they have been checked from the USPS scale because I use them to calibrate my bicycle power meter.

Attachment 1730130


Next, I applied 2 x 25lb masses to each end of a horizontal beam with end pulleys attached, connected via ropes (different color ropes to show that there is no 'sleight-of-hand'), with a centrally placed DILLON tension gauge (the same one I used to verify the 50 lb weights).

Photographic proof:

Attachment 1730131


Then read the DILLON tension gauge:

Photographic proof:

Attachment 1730132

And low and behold, the gauge read 50 lbf.

Exactly what static equilibrium states.

So, all is good, the earth still spins on its axis, the sun will shine tomorrow, William McCormack is a believer in science, and I do NOT need to get a refund on my engineering degree. Life is good!

The only problem is if the tension meter reads 50 there are 100 pounds of force acting on the cable. That is why ropes and cables and booms suddenly snap because you are working with twice the force that the labels on the weights and equipment report. In your mind when you lift 25 pounds it is no big deal, but your body is actually under 50 pounds of force to do so. The cable in the come-along when lifting 300 pounds is actually under 600 pounds of force which explains why it sounds like a guitar string being struck.

Sincerely,

William McCormick

Re: determined metal thickness needed

[/B][/QUOTE]

Originally Posted by SlowBlues

are the listed values weights against gravity or purely tension against fixed points?

I think the confusion lies in jumping between the two, force vs weight.

A 25-pound weight suspended by a cable creates 50 pounds of force on the cable the 25-pound force of gravity is in one direction and when you lift a 25-pound weight off the floor, it requires a counterforce of 25-pounds. The cable that can take an actual 50-pound force is sold as a 25-pound cable so that a weekend cable user can just weigh his payload and buy the cable that matches or exceeds that value. The hanging scale is calibrated to measure half the force it is being subjected to. The same as the platform scale is being subjected to twice the force of a weight it is measuring. A 25-pound weight free falling is being subjected to 25 pounds of force, to stop its freefall you must apply a counterforce of 25-pounds, which gives you 50-pounds of force total on whatever stops the weight. That is the reality of it, conventions change, not the reality.

Sincerely,

William McCormick

Re: determined metal thickness needed

All i know is, if you decide not to tighten the cap on an argon cylinder because it is cocked and on " Tight enough", then proceed to carry the cylinder by holding the cap....then you observe one of Newton's laws of motion. An object (cap) in motion will remain in motion until it meets an opposing force(chin) of equal or greater magnitude.

I also know that lifting blocks have reduced capacity when lifting vertically. Since the mass being lifted and the lifting force both act on the block.


Perhaps there is confusion of what is actually happening? I'on kno.


I still don't know what a trailing arm is.

Re: determined metal thickness needed

Originally Posted by William McCormick

Well if you agree that there is 180,000 pounds created by the two suspended weights then the two cables holding them are also under 180,000 pounds and the single cable holding both cables is being subjected to 360,000 pounds. Just like a hanging scale a cable holding a 25 pound weight off the ground against the force gravity and motionless is under 50 pounds of stress. The scale and the cable salesmen just sells them like they do to make it seem simpler for the new guy.

No, the upper cable has 180,000 lb, and each of the two lower cables has 90,000 lb - so combined, the two lower cables total 180,000lb. The sum of the vertical forces at the connection between the 2 lower cables to the 1 upper cable must = zero.

Originally Posted by William McCormick

The scale measures half the force upon it. A free falling weight with a cable and hanging scale attached is under 25 pounds of gravity force. When you grab the scale and apply 25 pounds of force against the force of gravity the scale registers 25 pounds but that is only measuring one of the forces acting upon it.

No, the scale does NOT measure half the force, it is measuring the only force that is applied to the system - 25 lb - that is in static equilibrium.


==> So you are not believing my math, and my graphics was not intuitive to you, so lets try this:


EXPERIMENTAL PROOF: The foundation upon which scientific theory is proven.



I used my DILLON tension gauge and a setup that mimics your 'scrap yard scale' pictorial:

First, measured the mass of the 2 x 25 lb weights to verify that they are indeed 50 lb total - measured in the vertical direction, using the DILLON gauge.

Photographic proof: A little hard to read in this photo, but it states 50 lbf. But I have other photos I can share if you need further proof. And the weights are stamped with 25lb - and they have been checked from the USPS scale because I use them to calibrate my bicycle power meter.

Attachment 1730130


Next, I applied 2 x 25lb masses to each end of a horizontal beam with end pulleys attached, connected via ropes (different color ropes to show that there is no 'sleight-of-hand'), with a centrally placed DILLON tension gauge (the same one I used to verify the 50 lb weights).

Photographic proof:

Attachment 1730131


Then read the DILLON tension gauge:

Photographic proof:

Attachment 1730132

And low and behold, the gauge read 50 lbf.

Exactly what static equilibrium states.

So, all is good, the earth still spins on its axis, the sun will shine tomorrow, William McCormack is a believer in science, and I do NOT need to get a refund on my engineering degree. Life is good!

Re: determined metal thickness needed

A cantilever by definition has to resist the forces levered against it, which depend on forces against and length of lever.

Every lever has a point of fulcrum, if lever length and forces on either end remain the same the force on the fulcrum point will not change if moved, only the ratios on either end. essentially the same as a cable split into two under tension with fixed points, gear/belt reduction, etc. all the same idea.

Most often the best way to simplify it is to look at both tension AND compression in each member. The "wall" of the cantilever doesn't need to support twice the weight, but it does need to resist two forces (compression and tension) in opposite directions (one side of base of wall pulling up, one side pushing down).

Re: determined metal thickness needed

are the listed values weights against gravity or purely tension against fixed points?

I think the confusion lies in jumping between the two, force vs weight.

Re: determined metal thickness needed

Originally Posted by Ingenuity

No, according to me (and science, physics, engineering, the real world, etc. etc.) - and the explanation below - the single upper cable attached to the dock crane will be acted upon by 180,000 lbf of force i.e. sum of the force in each of two container cables as the two are connected to the one dock crane cable.

Free body diagram of the upper pulleys and the 2-to-1 cable connection:

Attachment 1730113

Well if you agree that there is 180,000 pounds created by the two suspended weights then the two cables holding them are also under 180,000 pounds and the single cable holding both cables is being subjected to 360,000 pounds. Just like a hanging scale a cable holding a 25 pound weight off the ground against the force gravity and motionless is under 50 pounds of stress. The scale and the cable salesmen just sells them like they do to make it seem simpler for the new guy.

The scale measures half the force upon it. A free falling weight with a cable and hanging scale attached is under 25 pounds of gravity force. When you grab the scale and apply 25 pounds of force against the force of gravity the scale registers 25 pounds but that is only measuring one of the forces acting upon it.

Re: determined metal thickness needed

Originally Posted by Insaneride

Wacky Mac.

Loved your Mr Bill explanation. Saw him in post #17 also.
You are a pillar of welding world. You don't need stiñkingmath.
Keep up with your good work.

Edit: is it about tension or compression at the fulcrum tho? Whenever cantilevered.

A cantilever acts as a diving board. As the floor on the other side flexes it causes the cantilever angle to change. We fooled around with it while working at a new construction site. As weight was put on and removed from either side of the cantilever, there was an unnatural movement created on both sides. Many a time we would run into settling under the cantilever. It creates a dead blow hammer effect that pushes the foundation downward and can compress the plates holding up floor joists.

Sincerely,

William McCormick

Re: determined metal thickness needed

Originally Posted by Insaneride

Well now mister smarty pants, I do statistics in my head every single day so yes, I use math everyday. As a matter of fact for your information, at one point in time I used todo trigonometry in my head not only because I could but because I could and it saved time. So there

So we agree - we do need 'stinkingmath' :-)

Re: determined metal thickness needed

In a LINCOLN welding book, I read about structural design, no math. Simply put, you design a prototype , analyze failure and improve design. Math not required. Farmer way. No offense to farmers.
For fancy talkers: iterate design after failure, reiterate as needed. Hire Mathematicians for numbers and statistics for better results and profit. Bookkeepers and secretaries job.

Re: determined metal thickness needed

Originally Posted by Ingenuity

Well, I disagree. Whilst simple tasks of calculating lengths of weld is not multivariable calculus, it is math...and like it or not, you use it every day.

Well now mister smarty pants, I do statistics in my head every single day so yes, I use math everyday. As a matter of fact for your information, at one point in time I used todo trigonometry in my head not only because I could but because I could and it saved time. So there

Re: determined metal thickness needed

Originally Posted by Insaneride

You don't need stiñkingmath.

Well, I disagree. Whilst simple tasks of calculating lengths of weld is not multivariable calculus, it is math...and like it or not, you use it every day.

Re: determined metal thickness needed

Originally Posted by William McCormick

My graphic is depicting a dock crane lifting two 90,000 pound containers with one cable attached to two cables through a pulley system. According to you the two containers should only put 90,000 pounds of load on the dock crane.

Attachment 1730111

Sincerely,

William McCormick

No, according to me (and science, physics, engineering, the real world, etc. etc.) - and the explanation below - the single upper cable attached to the dock crane will be acted upon by 180,000 lbf of force i.e. sum of the force in each of two container cables as the two are connected to the one dock crane cable.

Free body diagram of the upper pulleys and the 2-to-1 cable connection:

Attachment 1730113

Re: determined metal thickness needed

Originally Posted by William McCormick

Yet the cable and scale are holding two 25 pound weights off the ground against 50 pounds of gravity. That according to my schooling requires 50 pounds of force to do. Or perhaps at the gym when you hold two 25 pound weights off the ground you are only holding 25-pounds against the force of gravity in your opinion?

A scale is calibrated to only recognize half the force upon it. A 25-pound weight in freefall represents 25 pounds of gravitational force, when you stop the 25-pound weight to weigh it, the scale applies a 25-pound counterforce to the object and records that it is applying a 25-pound counterforce to it.

Here we see Mr. Bill at the gym. Now physics requires that each one of Mr. Bills' arms must apply a little over 25 pounds to lift each 25-pound weight. After the weight is lifted it requires that a constant 25-pound force be maintained to hold the two weights off the ground. If Mr. Bill only used one arm to lift one weight applying 25 pounds of force he would be pulled over to the side of the weight machine unless he could use his feet, body weight, and stance to create a counterforce of 25 pounds.




You just cannot hold two 25-pound weights off the floor against gravity without 50 pounds of force. What you are trying to justify is that at the docks they could have one crane cable that splits into two cables over two seperate pulleys, and lift two containers using the same energy as lifting one.

Attachment 1730103

Sincerely,

William McCormick

Wacky Mac.

Loved your Mr Bill explanation. Saw him in post #17 also.
You are a pillar of welding world. You don't need stiñkingmath.
Keep up with your good work.

Edit: is it about tension or compression at the fulcrum tho? Whenever cantilevered.

Re: determined metal thickness needed

Originally Posted by Ingenuity

You are missing the pulley reactions to complete the total system. There are 2 pulley reactions (at opposing 45^o angles), and the sum of the vertical pulley components is 2 x 25 lb = 50 lb and that is equal to the magnitude of the 2 x 25 lb applied loads.

Neglecting any pulley friction (and mass of the rope) - a rope with an applied load of F, will have such a force (F) over is entire length. A pulley only changes the direction of the force, NOT its magnitude.

I can prove it to you with other some of my real-world work experiences, if you wish.

Or if you do not believe me, Google it. It is the internet - so it has to be true :-)

My graphic is depicting a dock crane lifting two 90,000 pound containers with one cable attached to two cables through a pulley system. According to you the two containers should only put 90,000 pounds of load on the dock crane.

Attachment 1730111

Sincerely,

William McCormick

Re: determined metal thickness needed

Originally Posted by William McCormick

Do you agree with this graphic? The Sprocket shaft fail is common.

Attachment 1730110

Sincerely,

William McCormick

Yes. Because 100 lb on the winch latch will be an equivalent force of 100 lb in magnitude, in a left-to-right horizontal direction. Sum of the forces in the horizontal direction (i.e. load applied to cable + force on latch) = force on sprocket shaft = 200 lb.

The force on the winch latch SHAFT will NOT be 100 lb - it will depend on internal dimensions of the latch and its bearing surfaces etc.

Re: determined metal thickness needed

Originally Posted by Ingenuity

Are you referring to this graphic you posted previously?

Attachment 1730099

If so, then yes, you are "missing something".

There is only 25lb of applied load at each end, and the tension in the rope/cable is also only 25 lb - so the scale also reads 25 lb.

Simple static free body diagram explains it. Cut a section through the horizontal rope and replace with unknown force, F:

Attachment 1730102

For static equilibrium:

Sum of the forces in the vertical direction must equal zero:

So: R_v = 25 lb.....[Eqn 1]

Sum of the forces in the horizontal direction must equal zero:

So: R_h = F.....[Eqn 2]

But, we also know that (assuming prefect pulley (frictionless) and massless rope) that the total pulley support reaction R_total is at a 45^o angle.

So R_total = R_v / cos(45^o).....[Eqn 3]

Knowing R_total we can calculate R_h = R_total * cos (45^o).....[Eqn 4]

Substituting Eqn 3 into Eqn 4 we have:

R_h = R_v /cos(45^o) * cos (45^o) = R_v

From Eqn 2:

R_h = F = R_v = 25 lb

==> Force in the scale = force in the horizontal rope/cable = 25 lb

Do you agree with this graphic? The Sprocket shaft fail is common.

Attachment 1730110

Sincerely,

William McCormick

Re: determined metal thickness needed

You are missing the pulley reactions to complete the total system. There are 2 pulley reactions (at opposing 45^o angles), and the sum of the vertical pulley components is 2 x 25 lb = 50 lb and that is equal to the magnitude of the 2 x 25 lb applied loads.

Neglecting any pulley friction (and mass of the rope) - a rope with an applied load of F, will have such a force (F) over is entire length. A pulley only changes the direction of the force, NOT its magnitude.

I can prove it to you with other some of my real-world work experiences, if you wish.

Or if you do not believe me, Google it. It is the internet - so it has to be true :-)

Re: determined metal thickness needed

Originally Posted by Ingenuity

Are you referring to this graphic you posted previously?

Attachment 1730099

If so, then yes, you are "missing something".

There is only 25lb of applied load at each end, and the tension in the rope/cable is also only 25 lb - so the scale also reads 25 lb.

Simple static free body diagram explains it. Cut a section through the horizontal rope and replace with unknown force, F:

Attachment 1730102

For static equilibrium:

Sum of the forces in the vertical direction must equal zero:

So: R_v = 25 lb.....[Eqn 1]

Sum of the forces in the horizontal direction must equal zero:

So: R_h = F.....[Eqn 2]

But, we also know that (assuming prefect pulley (frictionless) and massless rope) that the total pulley support reaction R_total is at a 45^o angle.

So R_total = R_v / cos(45^o).....[Eqn 3]

Knowing R_total we can calculate R_h = R_total * cos (45^o).....[Eqn 4]

Substituting Eqn 3 into Eqn 4 we have:

R_h = R_v /cos(45^o) * cos (45^o) = R_v

From Eqn 2:

R_h = F = R_v = 25 lb

==> Force in the scale = force in the horizontal rope/cable = 25 lb

Yet the cable and scale are holding two 25 pound weights off the ground against 50 pounds of gravity. That according to my schooling requires 50 pounds of force to do. Or perhaps at the gym when you hold two 25 pound weights off the ground you are only holding 25-pounds against the force of gravity in your opinion?

A scale is calibrated to only recognize half the force upon it. A 25-pound weight in freefall represents 25 pounds of gravitational force, when you stop the 25-pound weight to weigh it, the scale applies a 25-pound counterforce to the object and records that it is applying a 25-pound counterforce to it.

Here we see Mr. Bill at the gym. Now physics requires that each one of Mr. Bills' arms must apply a little over 25 pounds to lift each 25-pound weight. After the weight is lifted it requires that a constant 25-pound force be maintained to hold the two weights off the ground. If Mr. Bill only used one arm to lift one weight applying 25 pounds of force he would be pulled over to the side of the weight machine unless he could use his feet, body weight, and stance to create a counterforce of 25 pounds.

You just cannot hold two 25-pound weights off the floor against gravity without 50 pounds of force. What you are trying to justify is that at the docks they could have one crane cable that splits into two cables over two seperate pulleys, and lift two containers using the same energy as lifting one.

Attachment 1730103

Sincerely,

William McCormick

Re: determined metal thickness needed

Originally Posted by William McCormick

The scale in my graphic is showing that the scale is able to suspend 50 pounds against the force of gravity therefore it is under 50 pounds of force and registers half that force. Or I am missing something.

Sincerely,

William McCormick

Are you referring to this graphic you posted previously?

Attachment 1730099

If so, then yes, you are "missing something".

There is only 25lb of applied load at each end, and the tension in the rope/cable is also only 25 lb - so the scale also reads 25 lb.

Simple static free body diagram explains it. Cut a section through the horizontal rope and replace with unknown force, F:

Attachment 1730102

For static equilibrium:

Sum of the forces in the vertical direction must equal zero:

So: R_v = 25 lb.....[Eqn 1]

Sum of the forces in the horizontal direction must equal zero:

So: R_h = F.....[Eqn 2]

But, we also know that (assuming prefect pulley (frictionless) and massless rope) that the total pulley support reaction R_total is at a 45^o angle.

So R_total = R_v / cos(45^o).....[Eqn 3]

Knowing R_total we can calculate R_h = R_total * cos (45^o).....[Eqn 4]

Substituting Eqn 3 into Eqn 4 we have:

R_h = R_v /cos(45^o) * cos (45^o) = R_v

From Eqn 2:

R_h = F = R_v = 25 lb

==> Force in the scale = force in the horizontal rope/cable = 25 lb

Re: determined metal thickness needed

Originally Posted by Ingenuity

William:

That is incorrect. You only 'double the weight' if the cantilever span equals the backspan. For any other ratio of backspan/cantilever the distribution of support reaction is as follows:


Attachment 1730028

Attachment 1730029

This is first-year engineering statics - for some folks it was high school math/physics.




Evidently, it appears to be misunderstood by you. The scale is NOT calibrated to read half the load placed upon it. The scale reads the tension in the cable - and assuming frictionless pulleys - is equal to the load at one end. If the load at each end was NOT equal the loading system (in this example) would move (displace) until it was in equilibrium.

If you removed the left 25 lb pail (for example), and replaced it with a fixed support - like a clevis to secure the left end of the scale cable - and only have the 25 lb on the right end, the scale will read 25 lb. Action, reaction and equilibrium.

The scale in my graphic is showing that the scale is able to suspend 50 pounds against the force of gravity therefore it is under 50 pounds of force and registers half that force. Or I am missing something.

Sincerely,

William McCormick