Furcurequs (aka Dwayne) questions Jim_Mich

[Furcurequs]

jim_mich,

Thank you for showing your math. I believe I've now found the problem.

First of all, though, I would like to say that those of us with a 21.4 second time of acceleration to 50 RPM are in the right and you should have known that and not defended your incorrect answer since there are several different ways to do the calculation and it is also quite easy to do a quick check using a simple energy balance.

Now, with that said, why is your answer incorrect?

Well, it seems that the formula from your handbook for the radius of gyration of the compound pendulum may actually be wrong.

The moment of inertia of two point masses around the same axis of rotation is just the summation of the individual moments of inertia as you and/or others can see here:

http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html#mix

So,

I = m1*r1^2 + m2*r2^2

...and then the radius of gyration is just the square root of the (total) moment of inertia divided by the total mass as you and/or others can see here:

http://www.real-world-physics-problems. ... ation.html

k = (I/(m1+m2))^0.5

...or...

k = (( m1*r1^2 + m2*r2^2 )/(m1 + m2 ))^0.5

So, if I use this formula for the radius of gyration - which should be correct - instead of the formula from your handbook, then your method does indeed give an answer of 21.4 seconds too!

...which, of course, is the correct answer for a point mass approximation of the problem - which should be very close to the answer one would get even correcting for a real world geometry if the masses are reasonably compact.

I had actually thought that there was a possibility that your method would work, but I also knew there had to be something very wrong considering all the incorrect answers you were defending. That's why I wanted to see your math.

So, there it is.

That wasn't so hard, was it? ...lol

Take care.

Dwayne

[jim_mich]

So, what you're saying is that Machinery's Handbook is wrong? This book, which has been continually published and used for 100 years contains an error? And nobody has ever caught the mistake? And now, after all these many years, you say it's a mistake?

Back during the lifetime of Sir Issac Newton, many university professors, including Newton, believed that the formula for kinetic energy should to be kE = mass × velocity. Thus kinetic energy was thought to be linear to velocity. But the working class people who were designing machinery knew from hands-on experience that kinetic energy was exponential to velocity. Newton changed his formula to kE = m×v^2 primarily due to experiments published by Willem 's Gravesande. You remember Gravesande, he was the last person to witness Bessler's Kassel (4th) Wheel in operation on August 16th 1721, the day before Bessler destroyed it.

Gravesande published his kinetic energy experiments in a series of papers in Journal littéraire de la Haye, starting in 1722. In these experiments brass balls were dropped with varying velocity onto a soft clay surface. His results were that a ball with twice the velocity of another would leave an indentation four times as deep, that three times the velocity yielded nine times the depth, and so on. These results were shared with Émilie du Châtelet, who subsequently corrected Newton's kinetic energy formula from kE = m×v to kE = m×v^2. Note that though we now add a factor of 1/2 to this formula to make it work with coherent systems of units, the formula as expressed is correct if you choose units to fit it.

My point is that sometimes academicians get their formula wrong, while the lowly hands-on mechanic knows the correct formula. So in a situation where mass is spread out at different locations on a wheel or pendulum, is the correct formula that of the hands-on machinist as published for a hundred years, or is the correct formula that of the many learned professors?

It would be interesting to conduct some accurate physical experiments to see which is right, much like Gravesande did when there was originally some dispute about the KE formula. These two different formulas give results which are close enough alike that maybe the difference has been overlooked, being attributed to factors as friction or poor measurement.

Or can the correct formula be deduced in some logical manner proving the other formula incorrect?

[Furcurequs]

jim_mich,

I honestly do not know why the formula you used there for the radius of gyration does not give the value one would expect. It certainly does not seem to be a math error on your part for I get the same value that you do when I plug in the relevant numbers.

All I know is that the computer simulation, my equations, Wubbly's equations and a quick check by an energy balance all do agree - and that your method now, too, agrees and appears to work when I use the formula that I understand to be correct for the radius of gyration here.

I don't see any terms in your original handbook compound pendulum formula relating to the actual geometry of the masses or for the mass of a real world connecting rod, so I really don't know why the difference.

You can do a quick check yourself, though, of the original problem. You can easily calculate the speeds of all the masses when the (massless in this case) rod is spinning at 50 RPM. From that you can calculate the total kinetic energy of all the moving masses in the system - which is a lower value than you get using your handbook formula.

Using the formula for the radius of gyration that I had in my last post, it brings the value for the radius of gyration in this problem to 0.896 meters - and 2.5 kg moving around a circle of that radius at 50 RPM does give the correct energy value.

It's one of those head scratchers, I guess.

I do apologize for taunting you before. I needed to see for myself why your answer wasn't in agreement with all the other methods.

I'll probably do some searching online to see if I can find your handbook formula anywhere else.

Dwayne

[jim_mich]

Just to clarify...

The first portion of both of our formula's are the same: I = (m1 × r1^2 ) + (m2 × r2^2 )
Or can be written alternately as: I = (m1 × r1 × r1 ) + (m2 × r2 ×r2 )

But then Dwayne’s formula is: RofG = √ ( I ÷ (m1 + m2 ) )

While Jim’s Machinery’s Handbook formula is: RofG = I ÷ (m1 × r1 ) + (m2 × r2 )

They give similar but slightly different results.

This will take a little sleuthing to find out why the difference.

I might add that Machinery's Handbook was originally written a hundred years ago. At that time calculating the square root of a number by hand was very cumbersome. Sometimes alternate formula were devised that would give similar but not exact results. This could have been be the case here.

[Furcurequs]

Out of curiosity, I looked up information on the International Prototype Kilogram and it apparently is a cylinder of 90% platinum and 10% iridium (by mass) that has both a diameter and height of 39.17 mm.

http://greg.org/archive/2009/08/25/the_ ... and_k.html

So, if we were to steal two of the six copies of the one in the triple bell jar to use for our weights and then centered their axes on the ends of our meter long rods, what would be the contribution of their rotation about their own centers of mass in the problem we've been discussing?

Well, at the same rate of rotation, this energy contribution as a percentage is an additional (using the formula from one of my previous posts)...

100*1/2*(0.03917/2)^2

...or...

100*0.0001918

...or..

0.01918%

Which, of course, means that the contribution of the actual geometry for this concentrated a mass is indeed negligably small and that our point mass approximation for the probem is certainly sufficient to three significant digits.

For the sorts of gadgets we build, of course, with real world bearing friction and air resistance, such a difference would certainly be immeasurable.

Dwayne

[jim_mich]

When the ratio (weight size relative to radial distance) is small, as in the case you just sighted (1 to 51 ratio) then the RofG makes very little difference. But let's say you have a 4 inch diameter weight swinging on a 8 inch long lever. Then it becomes much more significant.

Now suppose you have 1/2 kg at 100 mm combined with 2 kg at 1000 mm. This gives you a combined object that is 900 mm across. Your calculations put the effective combined center of mass at about 35.26326 inches (0.89569 meter) while my Machinery's Handbook formula put it at 38.6414 inches (0.98149 meter).

What I'm saying is the Radius of Gyration formula makes the most difference when the combined weight-mass is spread way out.

[Furcurequs]

jim_mich,

Yes, I agree with your first statement. The relative proportions certainly do make a difference.

In regards to the other, I got my number from what I understand to be the very definition of the radius of gyration and when using that in your method it seems to be in total agreement with all the other methods for solving the problem as you can see below.

I honestly don't know what the deal is with your handbook formula, but it seems to give a value that is too large.

I believe you will find that computer simulations will bear this out.

The following number for the kinetic energy is what we would DEFINITELY have in the given problem when the rotation reaches 50 RPM.

Kinetic energy of two 1 kg masses rotating at 50 RPM at a 1 meter radius plus the kinetic energy of the 0.5 kg drive mass moving at 1/10th their speed downward:

KE = 1/2 * 2 kg * ( 2 * pi * 1 meter *50 RPM /60 (seconds/minute) )^2 + 1/2 * 0.5 kg * (2 * pi * 1 meter *50 RPM /60 (seconds/minute)/ 10 )^2

KE = 27.5 joules

Now, if we use my calculated value of the radius of gyration for the complete system we can calculate the kinetic energy thusly.

Kinetic energy of the total 2.5 Kg rotating at 50 RPM at a 0.896 meter radius of gyration:

KE = 1/2 * 2.5 kg * ( 2 * pi * 0.896 meter * 50 RPM / 60 (seconds/meter))^2

KE = 27.5 joules

So, the radius of gyration that I calculated gives the correct energy value and so then also the correct time and other stuff with your method.

You really can't dispute that this should be the kinetic energy value, either, because when we were told the geometry of the problem and the final rate of revolution - this kinetic energy value became a given.

I don't know what else to say. If you have different numbers - and a different energy value - then you have to try to explain where that energy is.

This is where I stand. Your method appears to work if the actual radius of gyration calculation is correct - but your handbook formula doesn't seem to give it.

Dwayne

[Furcurequs]

jim_mich

It seems to me that the the "l" given by your handbook formula may not be the radius of gyration about the pivot point at all but rather the length of the "equivalent simple pendulum" as is discussed here:

http://blog.cencophysics.com/2009/08/compound-pendulum/

They do appear to be different values.

Take care.

Dwayne

[ovyyus]

Jim's channel seems to have gone off-air with only static remaining. Yet another belief fails its champion?

[Trevor Lyn Whatford]

Hi Bill,

my signal is loud and clear, mind you I have been doing some fine tuning and checking the systems.

Joking apart, a lot of people here tie them selves up with one idea and narrow there field of vision, sceptics more than others, most think that physic is totally correct and based on empirically tested experiments when in fact the combination of mechanical arrangements mean that could never be the case, because like most here they stopped with the basics, when now things are just starting to get interesting, we are talking about thousands of untried experiments not just one or two, so in truth sceptics are only basing their Knowledge on theory not on facts absolute. In short there is nothing to stop a gravity wheel from working, it just needs the correct construction, so the fun is only just starting.

Jim I hope you can be a bit more opened minded and start looking at all the option available and put your intelligences to useful work, you use to come up with some clever designs so have a fresh start, anyway hope you are in your work shop where all the answer are.

[Furcurequs]

jim_mich

It seems to me that the the "l" given by your handbook formula may not be the radius of gyration about the pivot point at all but rather the length of the "equivalent simple pendulum" as is discussed here:

http://blog.cencophysics.com/2009/08/compound-pendulum/

They do appear to be different values.

Take care.

Dwayne

What I suspected is indeed the case.

Here is the formula that Jim used:



Link to Machinery Handbook at Google Books

...and here we find the explanation as to what the "l" is as described on the previous page in this edition of the Machinery Handbook:



Here are the relevant sentences:

The center of oscillation is the point at which, if all the matter in the compound pendulum were concentrated there, it would make a simple pendulum which would oscillate in the same periods of time.

...

In the formulas in the tables: l = the length of simple pendulum or distance between point of suspension and center of oscillation...

So, this means that the formula Jim used in his calculations does not give the radius of gyration but rather the length of the equivalent simple pendulum for a compound pendulum - and for the purpose of more easily determining the period of oscillation of the compound pendulum using the already familiar formula for calculating the period of a simple pendulum.

I posted this because some may still erroneously believe - like jim apparently believed - that he actually used the right formula from his handbook. That was not the case.

The actual formula for the radius of gyration of a two point mass system is simple to calculate and you really shouldn't need a handbook. One just adds the moments of inertia of the two point masses at their given radii and then divides that number by the total mass and then takes the square root of all of that.

So, in summary, had jim_mich used the correct formula for the radius of gyration in his calculations, his method would have indeed worked and given the same answer as we obtained using all the other methods.

Dwayne

[jim_mich]

A little bird told me that Dwayne is bashing me again. You just can't let it drop, can you?

Back in post post #121111 I explained why my answer was 25.5367 seconds, while all other peoples answers were between 21.358 and 21.4 seconds. I'll quote again:

The discrepancy is that with my calculations the weight masses out on the ends of the rod are fixed to the rods and thus are rotated, whereas the weights of WM2D were allowed to free-wheel and thus not rotate. Fixing the weights to the ends of the rods rather than letting them free-wheel results in it taking a slightly longer time to accelerate up to speed. In this case about 4.17 seconds longer.

In the above post, Dwayne failed to show the Radius of Gyration definition that is in Machinery's Handbook a few pages before the pendulum example.



And Wikipedia explains that Moment of Inertia is often defined in terms of Radius of Gyration...



In order to ACCURATELY determine the EXACT performance of any rotating or swinging object, one needs to know not only the radii to the center of the objects, but also the shape of the objects, and of course the mass of the objects.

Dwayne keeps harping...

The actual formula for the radius of gyration of a two point mass system is simple to calculate and you really shouldn't need a handbook. One just adds the moments of inertia of the two point masses at their given radii and then divides that number by the total mass and then takes the square root of all of that.

That is exactly what I did! But I went one step further to gain a little more accuracy.

Let me explain. A point-mass pendulum is fictitious. It can never exist. The reason is that when a pendulum swings, the bob at the end of the pendulum must rotate. The only time that a point-mass pendulum formula is accurate would be when the bob is attached so as to pivot at the end of the pendulum rod, thus allowing the bob to swing but not rotate. This is as close as one can come in the real world to a point mass pendulum.

Now if you have two weight masses, that are located at different locations upon a rod, and you are trying to calculate how this compound pendulum will swing, then in the real world such an arrangement can never be calculated by combining two point mass objects in the manner that Dwayne's formula would do. The reason is because being two separate masses, they will now swing around their pivot point of the rod, and at the same time the weight masses will rotate around their center of mass. Thus two weights don't act like a fictitious single point mass, except if you take into account the uniqueness of the particular situation.

The Radius of Gyration formula will give an ever so slightly different result than Dwayne's formula. The difference is caused by taking into consideration the rotating of the weight masses while swinging. Because the shape of each weight mass was unknown, I assumed each weight to be a point-mass weight object. But if you have two such point-mass objects, then when combined, you no longer have a point-mass. You have the equivalent of a solid object, which will swing around the rod's pivot point, and also the mass will rotate. The Radius of Gyration formula takes into account the rotation of the weight-mass. The two weight-masses are connected to each other by the rod. Thus they form the equivalent of one weight-mass which must rotate as it swings. This is what happens with a compound pendulum. Both weights swing, while at the same time the mass of both weights rotate. Dwayne's approach to solving the problem failed to take into account the rotation of the weight mass. Dwayne only calculated the swinging part. In the real life world, the mass must also be rotated. And this rotation requires energy. And as a result, in the real world, the oscillation velocity will be slightly slower. In the case of a wheel, the rotation speed will be slightly slower.

But of course I'm an idiot that doesn't know such things.

You guys can now go back to your bitching and back-biting and your sick sexual innuendos. I'll not participate.



Edit to add this link to post #22753 of a thread I started eight years ago title "Center of Gyration".

Second Edit: Link to start of said thread topic: http://www.besslerwheel.com/forum/viewtopic.php?t=1203

[murilo]

( wow!!! this guy is truly upset!!! B(((((

[Art]

.

Nice to see you're still around Jim ! ,

Now I don't have to keep wondering which new member might be a reincarnation of one of our most valuable members : )

.

[Furcurequs]

His value when it comes to his knowledge of basic physics certainly has much to be desired, however, if you have been paying any attention. After spouting that ignorant garbage he best be prepared for the storm that's about to be unleashed upon him.

...but for now he can enjoy the quiet before...

Dwayne

ETA: A little hint, though. jim_mich is the only person I'm aware of who has attempted this problem who hasn't actually been able to solve it. His first answer was off by 535% or so and his latest and closest to correct answer is still off by 19%. He's defended every mistaken attempt as if he alone new what he was doing and that everyone else was wrong. ...lol ...and, of course, now his current 19% error has to be due to the greater ACCURACY of his own unique methods. ...haha I don't have to "bash" him. I merely have to point out the sad truth about what he "believes" he knows. In other words, jim_mich simply discredits himself to those who aren't just plain ignorant.