Why only four pounds?

[daanopperman]

To finish ,

I think the weights in the wheel was more close to the axel than towards the rim , for I need less movement to bring the COM above the axel as if the weights was on the rim . It was said a small push only was needed to set the bidirectional wheel on its way . It must have been this small push that brought the COM above the axel . At standstill , the wheel would have been in a keeling state , the COM below the axel with the mechanism on or close to the rim .

[jim_mich]

Big wheel only lifted 112 lbs from axle. If wheel had 1100 lb. mass, why only 112 lb? That isn't logical, jim-mich.

IMO, the "Bessler PM Principle" was a rather weak effect relative the the amount of weight involved. IMO, it took a large amount of moving weight-mass to bring about a small amount of excess rotational force. Thus the machine had to be able to accelerate its own mass, while also providing the extra force to maintain its rotation and to do external work.

IMO, this is also one reason why Bessler's wheel has not been rediscovered. If a wheel puts out 120 Watts using 768 lbs of weight, then what might a smaller prototype wheel with say 32 lbs put out? Assuming a direct lineal proportion, it would be about 5 Watts (about 1/150 HP). This would very likely be totally consumed by friction. Thus someone building a small wheel containing small amount of weight, could very well stumble upon the Bessler design, but then reject it after testing as unworkable. And if it was a motion-wheel instead of a gravity-wheel, then few would even be seeking such a solution.

IMO, such a large heavy wheel that produced such little output was the main reason why Bessler had trouble selling his wheel.

Still it was a wonderful invention that proved PM possible. And if optimized it might be great and useful in some situations.

As usual, all are just my opinions.

[Dunesbury]

Yes, maybe so, but in my opinion wheel with that much mass would need more than one man to stop it.
How could one man stop 1000 lbs of spinning mass? Especially if he grabbed axle or handle rod with gloved hands, thus had less advantage?
The reason his PM principle hasn't been rediscovered is not because he was using large weight masses. Big wheel only maintained overbalancing force of about 3.2 lbs., i seem to remember. This could be accomplished with two, 4 lb. weights.

Total mass of wheel needs to logically agree with account that one man could stop it easily.

[jim_mich]

How could one man stop 1000 lbs of spinning mass? Especially if he grabbed axle or handle rod with gloved hands,

The Kassel wheel did not have any handle-rod on the axle. It did have the pulley-sticks arrangement for turning the water-screw. One might speculate that these were removable. From most accounts, it seems the Kassel wheel was stopped only by grabbing the rim. Stopping a 1000 lbs wheel rotating at the relatively slow speed of 26 RPM would not be all that hard.

As a reference, 26 RPM @ 12 foot diameter is 16.336 ft/sec, or 11.138 MPH (17.925 KPH). The 26 RPM is about 2.3 seconds per rotation. As a reference, the fastest human foot speed on record is 44.72 km/h (12.42m/s, 27.44 mph), seen during a 100 meters sprint, about 2.46 times faster than the wheel motion.

The question of course is what does it take to stop a 1000 lb, 26 RPM wheel? It may at first look, seem like more than human ability, but when you crunch the numbers, its not all that hard. Mostly because of the relatively slow 26 RPM speed of the wheel.

Also I might add that the 1000 lbs would NOT all be at the rim. It would be spread out. You might consider it centered somewhere between 50" and 54" radius. This would make the wheel act like about 725 lbs at the rim.

[jim_mich]

A 300 pound, 12 foot diameter solid disk rotating at 26 RPM should, then, have enough kinetic energy to lift the 200 pound person 3 feet 1 inch into the air, but for the person grabbing onto the rim where we have to consider the conservation of momentum, the person would only be lifted 1 foot 1 inch.

So might I assume a 1000 lb wheel, being 3.333 times as massive, should lift a 200 pound person about 43 inches into the air? I just checked, and my reach between grabbing something down low and grabbing it up high, without bending my body, is 50 inches. (Note that I have long arms.) So it would seem an ordinary man might be able to stop a 26 RPM 1000 lb wheel in one grab. Yes? No?

Does the man need to be actually lifted? Or is just the equivalent back-torque supplied by grabbing the wheel sufficient to do the same thing? I would say it makes no difference. If the force lifting the man is one ounce more than his weight, then he gets lifted. If one ounce less than his weight, then he stays footed on the ground. So there is no requirement that he be lifted. Only that he exert his whole body weight against stopping the wheel rotation.

[Fcdriver]

I'll let you know next week just exactly how hard it is to stop a wheel, and rather or not it lifts someone off the ground

[Furcurequs]

A 300 pound, 12 foot diameter solid disk rotating at 26 RPM should, then, have enough kinetic energy to lift the 200 pound person 3 feet 1 inch into the air, but for the person grabbing onto the rim where we have to consider the conservation of momentum, the person would only be lifted 1 foot 1 inch.

First of all let me say, oops, I should have shown my work on that one. It seems the second value there should have been more like 1 foot 4 inches instead of 1 foot 1 inch. I forgot to use the total mass in one of my momentum equations or something.

So might I assume a 1000 lb wheel, being 3.333 times as massive, should lift a 200 pound person about 43 inches into the air?

Well, if we are still talking a solid disk, then one that is 1000 lbs would have the same moment of inertia as just a 500 lb rim alone, so let's just use that since it makes things easier to calculate.

That would mean the stored kinetic energy in the 500 pound rim at 26 RPM would be:

ke = 1/2 * m * v^2 or

ke = 1/2 * 500 lbs / (32.17 lbs / slug) * ( (12 * pi ft / rot) * ( 26 rot/min ) * ( 1 min / 60 secs ) )^2

or 1/2 * 15.54 slugs * ( 16.34 ft /sec )^2

or 2074 foot*pounds

That is 2074 foot * pounds of kinetic energy, which is enough to get a 200 pound person about 2074 ft*lbs / 200 lbs or 10.37 ft or 10 feet 4 inches off the ground.

...which, of course, is 3 1/3 times the answer in my above example.

If we were to assume the person just suddenly latches onto the rim like I did in my example, in other words was able to just suddenly grab ahold of the side of the rim with his hands perhaps already fully extended above his head, we would caculate it as an inelastic collision.

So, the momentum of the rim before the "collision" would be equal to the momentum of the rim and the person after the collision and from that we can determine the "launch speed."

15.54 slugs * 16.34 ft /sec = ( 15.54 slugs + 200 lbs / (32.17 lbs / slug) * launch speed

So, our launch speed = 15.54 slugs * ( 16.34 ft / sec ) / (15.54 slugs + 6.22 slugs) or 11.7 ft / sec

After lift off, we now have 200 lbs of force decelerating the mass of the rim and the rider, so we can now calculate approximately how high he will go before coming to a stop.

v = a * t , s = 1/2 * a * t^2, s = 1/2 * v^2 / a ...where v is our launch speed

F = m * a , so a = F / m or 200 lbf / 21.76 slugs or 9.19 ft/sec^2

So, s = 1/2 * (11.7 ft/sec)^2 / (9.19 ft/sec^2) 7.44 ft or 7 feet 5 inches or 89 inches!!

By suddenly grabbing hold, our person will now be raised 89 inches into the air due to the much greater energy and momentum, so an assumption of 43 inches would be a bit on the low side, I would say.

So, wow, depending upon where his center of mass is, he could end up near the top of the wheel.

I just checked, and my reach between grabbing something down low and grabbing it up high, without bending my body, is 50 inches. (Note that I have long arms.) So it would seem an ordinary man might be able to stop a 26 RPM 1000 lb wheel in one grab. Yes? No?

This may actually be a more telling scenario than the one I gave. It would seem, though, that the answer to your yes or no question is a definite "no" in this case, however. ...or "no, no, no?" ...sorry...

Again, if we have the same geometry as in my problem with the solid disk, in your scenario with 3 1/3 times the mass, we've already seen that there is 3 1/3 times the kinetic energy in the wheel rotation.

Instead of treating the initial contact as a collision now, though, we would just assume the person is applying a force very close to his own body weight to the wheel and throughout a distance about equal to his reach. ...or throughout about 50 inches in your case.

Someone applying 200 lbs through about 4 1/6 feet will only take away about 833 ft * lbs of kinetic energy from the 2074 ft*lbs we calculated above, though. That means there would still be around 1241 ft * lbs of energy left in the wheel at the top of his reach! Ouch. That is essentially twice what I started with in my problem.

Someone would then have a decision to make. Would he let go or decide to go for a little ride?

If he decided to go on the ride, we would then treat the problem as an inelastic collision, since most of him is still not yet moving, and then use the new speed and momentum of the wheel at the very top of his reach in the calculations - where he would then effectively add his own mass to the mass of the rim during the "collision" and we could calculate how high he goes.

So, we need to find the speed of the rim after he's removed the 833 ft*lbs of energy.

1241 ft*lbs = 1/2 * 15.54 slugs * (new rim speed)^2

new rim speed = (2 * 1241 ft*lbs / 15.54 slugs)^0.5 = 12.64 ft/sec

Now we have to use that to determine the momentum before the "collision" (which finally gets our person moving) which equals the total momentum of all the mass after the collision and from which we can then determine our new "launch speed."

15.54 slugs * 12.64 ft/sec = (15.54 slugs + 6.22 slugs) * launch speed

launch speed = 9.03 ft/sec

Now, we can calculate the deceleration rate and determine how far he moves into the air.

v = a * t , s = 1/2 * a * t^2, s = 1/2 * v^2 / a ...where v is our new launch speed

F = m * a

a = F / m or 200 lbf / 21.76 slugs or 9.19 ft/sec^2

So, s = 1/2 * (9.03 ft/sec)^2 / (9.19 ft/sec^2) or 4.43 ft or 4 feet 5 inches or 53 inches!! Remember, this is now after you've tried to stop the wheel! Oh, yeah, and so we are movin' on up!!

Let's check that by calculating our total kinetic energy after the "collision" - the yanking of the person off the ground.

ke = 1/2 * 21.76 slugs * (9.03 ft/sec)^2 = 887 ft*lbs

...and then divide that by our rider's weight to get how high he is going...

height = 887 ft*lbs / 200 lbs = 4.43 ft Yep, it's the same answer.

So, this problem may actually be the barometer for determining about what kinetic energy and thus what moment of inertia the real wheel could have had during its operation.

Good thinking.

Approximately speaking, then, if the kinetic energy of the wheel is greater than the number we get by multiplying the person's body weight times the person's span of reach from low to high, then, the person will be lifted if he doesn't let go. ...this, of course, being the single grab scenario.

If, of course, he used poor form and just didn't apply enough braking force throughout his hold, he might still get lifted a little bit, however, even if the wheel numbers are lower.

If someone could stop the wheel with a single grab without actually being lifted, this scenario might indeed give us an approximate UPPER limit on the kinetic energy stored in the wheel's motion and so also its moment of inertia. All bets are off, however, if the person just applied pressure to the rim of the wheel until it came to a stop without an actual grip on the wheel or if he grabbed it multiple times, of course.

If a 200 pound person with your span of reach could stop the wheel without being lifted off the ground, then we can calculate the upper limit on the moment of inertia.

200 lbs * 4.167 ft = 833.3 ft*lbs = 1/2 * I * w^2 = 1/2 * I * ((2*pi rad/rot) * (26 rot/min) * ( 1 min / 60 sec))^2 = 1/2 * I * (2.72 rad/sec)^2

So, I = 2 * 833.3 ft*lbs / (2.72 rad/sec)^2 = 224.8 ft*lbs*sec^2/rad^2 (ETA: radians are dimensionless, so the units here are ft*lbf*sec^2 which is equivalent to slug*ft^2)

Since I = m * r^2 for all the mass at the rim, the maximum rim mass would be I / r^2

Maximum rim mass = 6.24 slugs or about a 201 pounds!

Let's emphasize that! If a 200 pound long armed man could with a single grab (and hold) stop the wheel without being lifted off the ground, the rim would have to have less mass than about 201 pounds.

Does the man need to be actually lifted? Or is just the equivalent back-torque supplied by grabbing the wheel sufficient to do the same thing? I would say it makes no difference. If the force lifting the man is one ounce more than his weight, then he gets lifted. If one ounce less than his weight, then he stays footed on the ground. So there is no requirement that he be lifted. Only that he exert his whole body weight against stopping the wheel rotation.

Well, as we can see from the calculations above, if the man can't take away all the kinetic energy from the wheel while he is still on the ground after an actual rim grab, if he doesn't let go of the rim, he will be lifted. Of course, though, multiple grabs or applying pressure to the rim could bring the wheel to a stop without having to worry about being lifted, anyway.

...but the evidence would seem to indicate a less massive wheel than perhaps you are thinking and also actually something more in line with my other calculations that you thought were suspect.

Dwayne

PS: I'll double check everything for errors later after I've had some time away from it.

[rasselasss]

To me,"The devil is in the detail",the wheel he "ran"for 53 days?...sealed in the room(without the "archimedes water feature) ,the bearings could'nt possibly survive with a 1000lb.load in the centre of a 1 inch axle,no matter how well wrapped with "fitted"wooden slats to make up the 8 inch axle ,the exposed 1 inch shaft ends each of at least 6 inches in length which would have been wrought iron (blister steel was not invented until 1740)with the axle bowed the axle ends would have serious "wobble"and the bearings would "run"in a few days......i would say a "realistic"total weight ..axle slats,wooden wheel,and weights could not safely exceed 550 pounds....the total weight "spread" no matter how well the slats are fitted is still applied at the bearings.....JMO.

[jim_mich]

Rasselasss, you obviously do not have an understanding of machinery. First, only half the weight, say 500 lbs, would be supported by each bearing. Second, I don't know where you get the idea of "axle bowed", because the wood part of the axle was constructed like a freight-wagon wheel, a bunch of wood staves held together by iron hoops. Such axles can carry massive loads. It would not "bow". Third, the axle pin was about 3/4 inch diameter iron, most likely turned on a lathe to a very smooth polished finish. The support bearing is assumed to be iron, or maybe bronze. It also would have been polished to a smooth finish. Such greased bearings can support very large loads, much more than 500 lbs.

Here is a link to a past thread where bearings were discussed.

[jim_mich]

Also, I personally find the 700lb estimate for the wheel a little high, as that seems like a lot for two men to translocate, even if you removed the weights; I lean more toward 400lb

An empty wheel weighing about 400 lbs seems about right to me also. If it contained 400 to 600 lbs of weights, then the weights needed to be removed before trans-locating. If the wheel contained 32 to 64 lb of weights, then the wheel could have been built lighter and more flimsy, say 340 lbs, and there would have been no need to remove the weights.

I'm not claiming that the wheel weighed 1000 lb when loaded. What I'm saying is that the upper possible limit of the weight of Bessler's wheel, when scrutinized, could be somewhere near 1000 lbs. There is no reason that says it could not be 1000 lbs.

[rasselasss]

Jim Mich,trust me i have an excellent understanding of machinery especially testing designs to destruction ..rotation machinery (Howden Sirocco)...if you read the post,i stated "the devil is in the detail"...we can't "see"or know for sure if this wood fabricated 8 inch shaft goes through the inside of the wheel, with the 3/4 or 1 inch shaft through the centre if this was so, your assumption may be correct...but if you look at the drive spikes on the wheel shaft which turn the rope to turn the archimedes water feature ...their appearance is clear through the axle centre....they would HAVE to be or they'd shear off...as i said before these decrepancies like the scale drawing (which suggest bush bearings,and the length of landing area for the shaft in the support pillars)most everything is purely assumptions...i have to go with lightweight not heavy....lets get real here 53 days no grease applied in that time ....has to be lightweight..

[rasselasss]

Big wheel only lifted 112 lbs from axle. If wheel had 1100 lb. mass, why only 112 lb? That isn't logical, jim-mich.

IMO, the "Bessler PM Principle" was a rather weak effect relative the the amount of weight involved. IMO, it took a large amount of moving weight-mass to bring about a small amount of excess rotational force. Thus the machine had to be able to accelerate its own mass, while also providing the extra force to maintain its rotation and to do external work.

IMO, this is also one reason why Bessler's wheel has not been rediscovered. If a wheel puts out 120 Watts using 768 lbs of weight, then what might a smaller prototype wheel with say 32 lbs put out? Assuming a direct lineal proportion, it would be about 5 Watts (about 1/150 HP). This would very likely be totally consumed by friction. Thus someone building a small wheel containing small amount of weight, could very well stumble upon the Bessler design, but then reject it after testing as unworkable. And if it was a motion-wheel instead of a gravity-wheel, then few would even be seeking such a solution.

IMO, such a large heavy wheel that produced such little output was the main reason why Bessler had trouble selling his wheel.

Still it was a wonderful invention that proved PM possible. And if optimized it might be great and useful in some situations.

As usual, all are just my opinions........

....I would say this is an accurate statement concerning the "power"of the wheel......lightweight,low power..

[Furcurequs]

"An attempt to stop it suddenly would carry a man to near the ceiling."

;P

Okay, here's Joseph Fischer's actual (translated) quote:

An attempt to stop it suddenly would raise a man from the ground.

...and here it is with the prior sentence:

I then stopped the wheel with much difficulty, holding on to the circumference with both hands. An attempt to stop it suddenly would raise a man from the ground.

If one could interpret these words to mean that Fischer could actually stop the wheel with but a single grab and hold without being lifted from the ground, this would give us an idea as to the upper limit of the wheel's rotational inertia. ...and it would, as per my prior calculations, probably be something a bit less than that of just a 200 lbm rim alone.

When Gravesande said the wheel was "very light," he probably meant it.

Of course, mass nearer the axle doesn't have much of an effect on the rotational inertia. This would simply imply that if it is more massive, there is not so much mass near the rim.

Dwayne

ETA: Just for the heck of it, I calculated that a 200 lbm rim would have about the same moment of inertia as that of a 100 lbm rim with a near 6400 lbm 2.1 foot diameter depleted uranium core. :P

[rasselasss]

At best i have to take the scaled drawing by (can't make out the name )because of discrepences in relation to the witness statements (uncapped bearing examinations ,the stopping the wheel by hand etc.) as artistic licence value only....the drawing shows a "stopping"handle? (c) in board of the (b)"drive spikes(which also appear to go through the axle centre)"beside the wheel flange through the axle centre (splitting the iron axle within which would not be sound engineering practice ).....i honestly find it strange important details are vague not only in the drawing but the lack of important details (wheel weight etc.)....from the witness's...

[rasselasss]

http://www.bing.com/images/search?q=bes ... tedIndex=0