energy producing experiments

[Furcurequs]

Hey pequaide,

I don't know if you saw my last post directed to you in the other thread or not. It was covered up pretty quickly by other stuff, but anyway, in that post I addressed the Smokin' Lamas and analyzed a bit of data from a youtube video of one of the pumpkin tosses.

In that particular one it appears the launch speed was about 176 ft/sec or close to 54 m/sec when the wheel was spinning at around 360 rpm. I compared the interval traveled between frames with the wheel diameter which I assumed to be about 6 feet, so the calculations are only approximate. Maybe you have more information on the actual dimensions?

Anyway, I'll just go ahead and quote that post here so others can at least see from my gif animations the device you are speaking of.

pequaide,

Before I get back to addressing jim_mich's mistaken math, which he still seems to be in denial about to some degree, I thought I would just go ahead and address your stuff since I already had most of this post written.

In the thread I spoke of in which jim_mich had that mistaken math, you seem to have also been mistaken in yours. We know why yours is wrong, though, as I and apparently several others have tried to explain to you before. You leave one of "r"'s out of the moment of inertia equations - while apparently just choosing not to accept what are accepted fundamentals of physics.

Now, with that said, to address one of your questions about the Smokin' Lamas:

We know that the distance of 1776 feet is the horizontal distance that the pumpkin traveled while it was in the air. During the travel time, though, it also both ascended and then descended in the vertical.

So, to get a rough ball park figure for the minimum launch speed, we can easily do some calculations assuming no air resistance and that the pumpkin was launched from the same vertical level as that on which it landed. For maximum range with no air resistance we would have also wanted to have launched at a 45 degree angle, too.

That means that the magnitude of the launch velocity (v) multiplied by the sine of 45 degrees will give us the magnitude of the velocity in the vertical and the magnitude of the launch velocity (v) multiplied by the cosine of 45 degrees will give us the magnitude of the velocity in the horizontal.

We know that the velocity in the vertical is slowed to a stop during the ascent by the acceleration due to gravity (g) and then the acceleration due to gravity (g) then speeds it back up in the opposite direction. Since "g" is a constant, we know that the time of ascent must also equal the time of descent (again, assuming no air resistance ) and that the final speed in the vertical will be the same as the initial speed but in the opposite direction.

Let's now set up our equations:

v * sin(45degrees) - 32.2 feet/second^2 * t = 0 feet/second

In the above equation "v" is the magnitude of the launch velocity and "t" is the time of ascent - or the time it takes the pumpkin to come to a stop in the vertical direction.

We now have one equation and two unknowns, so we need another equation:

v * cos(45 degrees) * (2 * t) = 1776 ft

In this second equation "v" is again the magnitude of our launch velocity and "t" is again the time of ascent, so the speed in the horizontal multiplied by the total time in the air equals our horizontal distance traveled and so I'm multiplying "t" by 2 in this equation since the time of ascent and descent are equal.

Now we can solve for the initial velocity "v" (and leaving out some gory details):

v = ( ( 1776 feet * 32.2 feet / second^2 ) / (2 * sin ( 45 degrees) * cos ( 45 degrees ) ) )^0.5

v = 239 feet/second or 163 miles/hour

t = 5.25 seconds , so total time in the air is about 10.5 seconds

Anyway, that should be a pretty good ball park figure for a minimum launch speed. The actual launch speed would, of course, have to have been greater due to air resistance losses. Also, one might want to consider and possibly correct for differences in the vertical height of the launch and the landing positions if one had more data.

I actually went seeking some data and found this nice video of a side view of a Smokin' Lamas pumpkin toss:

http://www.youtube.com/watch?v=CsjzJKKhMlU

I extracted the videos frames of just the launch and put them in this slow motion animated gif:



You can see that the pumpkin is quite clear and that there seems to have been very little motion blur. So, I decided I'd just overlay the pumpkin from several different frames onto one and got this:



While the pumpkin is in the air the camera also seems to be pretty stable, too, so that was quite fortunate.

Here is the same thing but with the sling from the previous three frames also overlaid, but unfortunately I've noticed that it appears there may actually be a missing - as in a "dropped" - frame from the video right around the actual time of release. There is more than twice the distance between the first view of the pumpkin in the air and the last view of the pumpkin in the sling than between any other two shots of the pumpkin in consecutive frames, so unless at the release the pumpkin lost over half its speed and over 3/4ths its kinetic energy - which almost certainly wouldn't have happened - there is indeed a problem with the video.



Anyway, if we just look at the intervals between pumpkins in the air, though, we can see that the first apparently reliable interval is very close to the same distance as the diameter of the wheel. Guessing at around a 6 foot diameter for the wheel (after having seen pictures of people standing next to it) and using the frame rate of 29.33 frames per second, that would put the speed of the pumpkin at about 6 feet * 29.33/second or approximately 176 feet per second which works out to be about 120 miles/hour.

In this animated gif I made from frames taken from immediately before the launch you can see that the wheel turns just slightly more than a full turn in 5 video frames.



Here is an animated gif using only the zeroth and fifth frame from the above and so if you look at the end of the red spiral you can see just how close to the same position they are. I chose these frames so you can use the spot of sunlight and/or the line there in the background as a reference.



Anyway, this would indicate that the speed of the wheel is around 360 rpm.

It's hard for me to find specific details, but I did find on the Smokin' Lamas Facebook page that the latest wheel (which I'm not really sure that this is, btw) weighs in at 665 pounds.

If the wheel in this video is the 665 pound version and the mass were distributed uniformly as if in a solid cylinder, then at this rotational speed the wheel should have had over 10 times the energy that the flying pumpkin ends up with. If the mass were mostly in the rim, it would have been over 20 times the energy of the pumpkin.

So, most of the wheel mass may actually be concentrated near the center here.

You said the tether length doesn't matter, but by drawing a line through the pumpkins shown in the air and then drawing a line through the center of the wheel that is at right angles to that, we can determine the moment arm and calculate the angular momentum of the pumpkin around the center of the wheel.

I wish I had more details about the wheel itself, like the actual mass distribution and where the pumpkin rides in the wheel before it is launched, so that I could properly model the whole thing, but one might be able to get approximations for those things using equations based upon the conservation of energy and the conservation of angular momentum.

After thinking about it, if these folks knew their physics they could probably reduce the wheel mass significantly and get the same results. With a (2x) bicycle wheel type design with most of the mass near the rims, they could maybe reduce the weight to nearly 1/10th of what they have. It would be a lot easier to transport and mount, I would think.

Of course, though, without really knowing specifics about what they really do have, I'm just sort of speculating. There may also be some strength considerations that come in to play, too.

Oh, also, from reading the rules of the main punkin chunkin contest that they enter, it seems the person pedaling their wheel up to speed only has a maximum of two minutes in which to do that.

From what I've read, weightlifters can for very brief periods put out nearly 3 horsepower in an anaerobic effort. More sustained aerobic efforts like that of cyclists, though, max out at only a little over half a horsepower.

Highly trained cyclists can supposedly output around 400 watts for as long as an hour. "Highly trained," though, in cycling probably also includes the taking of illegal performance enhancing drugs and blood doping, if you keep up with such things. ...haha

Also, with actual cycling the body is cooled from the motion through the air which ups the sustainable power output too, since the body itself is essentially a heat engine that works more efficiently when kept cool.

Anyway, if the guy doing the pedaling was putting out about 1/3 horsepower for 2 minutes to bring the wheel up to speed, that would be about 3 times the energy the pumpkin has here - if my quick calculations were correct.

In other words, they are not getting something for nothing.

Also (and as even with Bessler's wheel) it would be rather nice to have some data on the air resistance of an actual spinning disk like that to know how much energy was wasted just getting the thing up to speed.

Pequaide, I do like your experiments and I think it would probably be a good thing to better understand how quickly and efficiently energy from a flywheel can be transferred to a projectile using a tether like that, but still, there seems to be no sign of any excess energy in these experiments and they are quite well modeled and explained using traditional concepts and math. ...if, of course, one has all the correct parameters.

BTW, I just put out over 1/2 a horsepower myself doing A chin-up! I'd need to do 120 in two minutes, though, to be able to sustain that. Maybe I should move to jumping jacks. ...or squat thrusts?

Take care.

Dwayne

http://www.besslerwheel.com/forum/viewt ... 039#118039

Dwayne

[pequaide]

There is a cage that caries the pumpkin near the center of the wheel. I would guess that the 665 lbs includes the axle.

The quick reduction in the distance between the photographed pumpkin is consistent with other electronic data. There is a very quick reduction in velocity.

I went from a video verbal for the RPM and it was much lower than what you estimated.

Also note that the throwing angle is very shallow. This is done because of the quick reduction in velocity. You have to get it out there before it loses its motion.

Also I see a backward motion in the wheel as it throws. Any motion in the wheel is lost motion.

Plus look at the frame as it throws. The frame lifts as the wheel throws. If the wheel is 665 lbs the frame is surely well over a ton.

[pequaide]

Under ideal condition the universe chooses to create binary systems that have equal momentum on both side of the common center of mass. From stars and planets to baseball bats flipped in the air.

So why wouldn’t one think that smokin lamas is trying to create a binary system that has equal momentum on both side of the common center of mass between the pumpkin and the mass of the wheel. Every other free moving object in the universe finds this center of linear Newtonian momentum; why wouldn’t smokin lamas? But from the motion of the frame of smokin lamas it appears that the common center of mass has the frame on the other side as well.

You can prevent this balanced binary system from being found by the placement of bearing and levers etc. but the spinning system will be trying to find it.

From my calculation the motion of the frame of smokin lamas contains about half of the motion of the system. This calculation was made independently of the fact that this is what the system is trying to do; which is to give half the motion to the other side. There are ways to isolate the pumpkin’s motion away from the frame, but I will not review it right now.

And the air resistance of the pumpkin does not begin after release from the sling, air resistance exists before release.  As soon as the pumpkin and sling bag leaves their position near the center, massive amounts of air resistance are placed upon the sack as the velocity builds. The time covered by the three frames before sling release probably have similar drag to that drag found in the three frames after the pumpkin released from the sling.  The sling bag is accelerating in the time before pumpkin release but that does not mean there is no air resistance.

If the Earth and Moon were placed at rest in deep space they would accelerate toward each other and they would constantly be acquiring the same momentum. As they approached each other they would have the same linear Newtonian momentum.  Gravity can only give them the same linear momentum; gravity cannot give them equal energy or equal angular momentum it can only give them equal linear Newtonian momentum. So it should be no surprise that equal linear Newtonian momentum is found in the Barycentric orbit.

But why does the equal and opposite gravitational force find it just as easy to hold the same quantity of linear Newtonian momentum in orbit on both side of the center of mass after the barycenter system is formed? Could it be that gravity not only causes Linear Newtonian momentum but that gravity also maintains it.

A 2 kilogram bullet fired sideways to the Earth’s preexisting motion will have to have a velocity change of 1000 m/sec to deliver 1,000,000 joules of energy. And a bullet fired with the motion will only have to have an increase in velocity of 17 m/sec to give you 1,000,000 joules.  Further; all these rotational motions that cause the preexisting motion of the Earth (Milky Way, solar system, axis) would be constantly adding and subtracting from each other.

If the energy of the universe were constant; a u-boat captain would have to know if he was firing his deck cannon with or sideways to the earth’s motion.  A discharge with the motion would not clear the length of the sub. A firing sideways to the earth’s preexisting motion would be normal.  The u-boat captain would have a complex set of calculation just to hit a merchant ship 300 meter away. We know that the motion of the earth is real and we know how to calculate kinetic energy; but why doesn’t the kinetic energy made remain constant and equal with the chemical energy expended by the shell.

The u-boat captain can fire in any direction he wants because the Law of Conservation of Energy is of course false. The Law of Conservation of Energy has somehow become the sacred cow of science.

The shell’s gun powder does however produce a uniform quantity of linear Newtonian momentum that is independent of the direction in which it is fired. 

[pequaide]

From Wikipedia: The magnitude of the centripetal force on an object of mass m moving at tangential speed v along a path with radius of curvature r is:[5]
F = mv²/r

Then why does Wikipedia in the 'despin device' post say this of the small weights; "The relatively small weights can have such a large effect since they are far from the axis of the spin, and their effect grows as the square of the length of the cables."

They seem to be applying a formula that has no application to the actual event.

The only effect is the pulling of the cable to get the satellite to stop rotating; and that pull is centripetal force.

The force in the cable is centripetal and it will decrease as the cable (r) lengthens.

The only way to get an appropriate effect is to have a massive increase in the tangential velocity. And this occurs when the cable is still short.

[pequaide]

http://video.mit.edu/watch/mit-physics- ... hine-3138/

Yes: the 10 gram is causing all the motion; it is the only unbalanced force. This is 10 grams accelerating 1110 grams
.
I did not say their experiment made energy. But add to this the experimental concept proven by the double and triple Atwood's and you have a means of making energy. This concept is that the Laws of Levers applies to Atwood's pulleys connected over the same axis but with different radii. As described below.

Let’s say that the radius of the pulley is 8cm.

550 grams on the left side at 8 cm places a certain torque on the point of rotation.  This torque is perfectly countered with an equal amount of torque from 550 grams on the right side at 8 cm.

The extra 10 grams on the right side is the accelerating force of .01 kg * 9.81 N/kg = .0981N which is expressed as torque upon the center point of rotation. These 10 grams of measured force remains constant throughout the acceleration. There is not one force for the static position and then another force for the acceleration. Acceleration can be determined from F = ma; F/m = a; .0981 N / 1.110 kg = .08838 m/sec²

The individual torque of each 550 grams was caused by the gravitational pull on the mass. The one side’s torque canceled the others side’s torque but the inertia of the 550 (1100 grams) grams is not canceled.  This inertia is proportional to the previously existing torque.

In Newtonian Physics inertia is measured by the change in momentum (linear Newtonian momentum) caused by an applied force. The change in linear Newtonian momentum of both 550 gram sides is equal. The bobs are viewed as going in the same positive direction.

55 grams at 80 cm has an equal amount of torque as 550 grams at 8 cm. If you try to move the 55 grams at 80 cm you will find that it acts like it has an equal amount of inertia from the point of rotation as 550 grams at 8 cm.  When moved; it also has an equal amount of linear Newtonian momentum, and an equal amount of centrifugal force. The system is balanced and it acts like it.

The 10 gram will accelerate the 55 g at 80 cm just as easily as 550 g at 8 cm; in fact it will accelerate two 55 grams on each side at 80 cm just as easily as it accelerates two 550 grams at 8 cm on both sides. The final velocity for the 10 grams (left at the 8 cm position) with two 55 grams at 80 cm will be exactly the same as when you had two 550 grams at 8 cm; with the additional 10 grams at the 8 cm position. And the velocity of the 10 grams will be .42 m/sec. Velocity can be determined by a rearrangement of the distance formula: v = the square root of (2 * d * a) = the square root of (2 * 1 m * .08837m/sec²)

After a drop of one meter for the 10 grams; the 110 (55 + 55) grams on a 80 cm wheel will be moving 4.2 m/sec; this (1/2mv²) is .971 joules of energy. The input energy was (10 grams dropped one meter.01 kg * 9.81 N/kg *1 m =) .0981 joules. Roughly 10 times the energy
For a logical proof place 550 grams on one side at 8 cm and 54 grams on the other side at 80 cm. The 550 grams will rotate the 54 grams.  Place 550 grams on one side at 8cm and 56 grams on the other side at 80 cm. The 550 grams will be rotated by the 56 grams. With 55 gram on the other side at 80 cm there will be no rotation.

Now place 550 grams on one side at 8cm and 549 grams on the other side at 8 cm. The 550 grams will rotate the 549 grams.  Place 550 on one side at 8 cm and 551 grams on the other side at 8 cm. The 550 grams will be rotated by the 551 grams.  With 550 grams on both sides there will be no rotation.

Conclusion: the Difficulty of rotating 55 grams at 80 cm is somewhere between the difficulty of rotating 549 grams at 8 cm and 551 grams at 8 cm.

When an outside force is asked to accelerate 55 grams at 80 cm it will find it as difficult (somewhere between 549 grams at 8 cm and 551 grams at 8 cm) as rotating 550 grams at 8 cm.  Don’t buy into the myth that a different force takes over as soon as motion begins. Their experiment proves that the 10 grams of force starts the motion and it stays with the system throughout the one meter drop. The 4.80 seconds is predicted for the 10 grams of force (.0981 N) by F = ma. d = 1/2at² = 4.757 seconds.
The 110 grams moving 4.2 m/sec could rise .889 meters for .110 kg * 9.81 N/kg  * .889 m = .9702 joules

Detractors will try to angular momentum you; but Atwood’s are linear and Angular Momentum is for space.

[pequaide]

At some point the opposition is going to have to defend an imbalanced wheel. Their mr² theory only puts 5.5 grams at 80 cm. At low RPM the bearing could withstand the centrifugal imbalance from one side to the other. But at high speed the centrifugal force difference would build and the bearing would be ripped to pieces. This is because the centrifugal force being given to one side (as the RPM builds) is much greater than the centrifugal force being given to the other side. And this centrifugal force difference will be used to destroy the bearing. This would be the absolute value of the force not the percentage.

This is some definition of balanced; a wheel that tears itself apart.

In the 550 gram at 8 cm on one side and 55 grams at 80cm on the other side there would be no centrifugal force difference from side to side; no matter what the RPM. Even a light duty bearing would have a service life of several years.

[rlortie]

To all those who own vehicles, especially those who own heavier models and trucks. Park them and walk, all the force from the vehicle weight is transferred to the bottom half of your axle bearings even when stopped, your bearings are about to be ripped to pieces!

This is shocking to hear, the wheel bearings in my 3/4 ton Dodge Ram with a gross weight of over 7,000 pounds has traveled over 130,000 miles. I fear that they may explode on me any day now! :-)

Residents of countries that drive on the left side of the road should take extra precaution, exploding bearings may cause you to end up in the right lane facing oncoming traffic. Remember, it only takes 5.5 grams at 80 cm at moderate speed to cause your bearings to explode.

This is true as I read it right here!

[pequaide]

Please stay off of the thread rlortie. thank you

[rlortie]

OK pequaide, I promise I will stay off your thread, you do not need my assistance in making a mockery of yourself.

Regards,

Ralph

[pequaide]

I found one double radius Atwood's pulley on the net. They appeared to find acceleration the same way I would. Which means they are trading mass for speed, but getting the same momentum. So someone agrees.

I won't tell you the site; I am sure they don't need the abuse.

[pequaide]

Why is it that you push for ten times as long and get the same quantity of motion?

This experiment pulls for over ten times as long but gives you the same quantity of motion.

The same 10 grams of force pulls for ten times as long under the nearly frictionless Atwood's system and it gives you the same kinetic energy as 10 grams free falling the same distance.

The free fall of the ten grams takes .4515 seconds to fall one meter and it gives you a final velocity of 4.429 m/sec for .09808 joules of energy.

The final velocity for the Atwood's is .4204 m/sec for the 1.110 grams and that is also .9808 joules of energy. And the Atwood's takes 4.757 seconds.

But: is it logical that 4.757 - .4515 = 4.30 seconds of pulling should come to nothing?

Some sight the energy equivalence for free fall and the Atwood's as a victory. Or is it an embarrassment; to pull in a friction free experiment and have 4.3 second of pulling come to nothing.

The free fall momentum of the ten grams is .04429 units. The final momentum of the Atwood's is .4666 units. This is 10.535 times as much momentum. The time difference is 4.757 sec / .4515 sec = 10.536. This is with the force remaining the same in the two friction free systems; an Atwood's and free fall.

Second Question: Shouldn't it take the same amount of energy to stop an object as it did to start the object in motion in the first place?

Just as the Atwood's reaches the end of the experiment at the meter point; you could have the up going (550 g) mass combine with a 30 gram mass at rest. Ten of those 30 grams would be used to create a new balanced system, and that would give you 20 grams left over. You now have 560 grams on one side and 580 grams on the other. The twenty grams would decelerate the Atwood's and make it come to a stop. Lets see how much 'Force times time' it takes and then how much 'kinetic energy' it takes to make it stop.

The original momentum of the Atwood's is .4666 units and that is divided between 1.110 kg + .030 kg = 1.140 kg for a new velocity of .4089 m/sec the new acceleration is 20 g/ 1,140 g * 9.81 m/sec² = .1721 m/sec². So this is a new stopping distance of .4857 meters, and a new stopping time of 2.3757 seconds. There is twice the force so the new Ft = mv is 2.3757 times twice the force ( 20g or .1962 N) which is the same as .0981 N for 4.75 seconds. Ft is the same to stop the motion of the Atwood's.

But for the 30 grams at rest added to the upcoming 550 grams you need only 97.1 % of the energy to stop it. .48576 m *.1962 N =.0953 / .0981 97.1 % of the original energy.

With the up coming mass picking up 60 gram the energy loss is down to .0930 joules remaining but that is all that is needed to stop the Atwood's. This is a new acceleration and distance but it is only 94.8% of the original energy. But under the Ft rules 100% of the same Ft is needed to stop the Atwood's

And after the one meter drop the upcoming mass of the Atwood's picks up 110 grams the energy needed to stop the Atwood's is only 91% of the original energy. The Ft is always the same as the original.

And after the one meter drop the upcoming mass of the Atwood's picks up 410 grams the energy needed to stop the Atwood's is only 73% of the original energy. The Ft is always the same as the original.

So the original quantity of energy is stopped by 97.1%, 94.8%, 91% and 73%.

Kinetic energy is a useful tool; but it is not a conserved quantity. How can you stop something using only 73% of the energy that was needed to make the motion in the first place?

Both these concepts ( mv and mv²) were contemporary with Bessler; could he have used the difference between the two concepts to arrange a pendulum trebuchet as in Alex's recent post?

[pequaide]

There is current research with Yo-yo despin devices, they list tether lengths that would leave the thrown mass with less linear velocity than the spin of the satellite, if angular momentum was conserved. Is that possible?

[pequaide]

The satellites have extremely long tethers. When the tethers become longer, as the small mass unwinds, the radius increases and that would mean that the linear velocity of the thrown mass must continually decrease for angular momentum conservation. At some point the thrown mass must actually have a smaller linear velocity than it originally had. Does that mean that the spinning motion of the rest of the mass of the satellite disappeared?

What would happen if the small mass were to return its motion to the satellite? Can one kilogram moving five meter per second cause 1000 kilograms to move 6 meter per second. Do the math the extremely long tether would require the impossible. Angular momentum conservation can not occur unless there is gravitational acceleration or deceleration, and yo-yo despins have no gravitational acceleration and therefore do not conserve angular momentum as NASA claims. They never plug in numbers to their massive formulas. And there is a good reason; they don't work.

[Mark]

There is current research with Yo-yo despin devices, they list tether lengths that would leave the thrown mass with less linear velocity than the spin of the satellite, if angular momentum was conserved. Is that possible?

The satellites have extremely long tethers. When the tethers become longer, as the small mass unwinds, the radius increases and that would mean that the linear velocity of the thrown mass must continually decrease for angular momentum conservation. At some point the thrown mass must actually have a smaller linear velocity than it originally had. Does that mean that the spinning motion of the rest of the mass of the satellite disappeared?

What would happen if the small mass were to return its motion to the satellite? Can one kilogram moving five meter per second cause 1000 kilograms to move 6 meter per second. Do the math the extremely long tether would require the impossible. Angular momentum conservation can not occur unless there is gravitational acceleration or deceleration, and yo-yo despins have no gravitational acceleration and therefore do not conserve angular momentum as NASA claims. They never plug in numbers to their massive formulas. And there is a good reason; they don't work.

Current research ??

Yo-yo de-spin was originally invented, built, and tested at Caltech's Jet Propulsion Laboratory.
http://en.wikipedia.org/wiki/Yo-yo_de-spin

They filed for patent in 1959.
https://www.google.com/patents/US3030049

This technology has been used successfully for decades.

Even with a 204-kilogram (450-pound) third stage (which was 2224 kilograms, or 4903 pounds, before it began expending its propellant) and a 1218-kilogram (2685-pound) spacecraft, the small yo-yo system halts the spin and even reverses it, leaving Dawn rotating at 3 rpm in the opposite direction from its original spin. About 1 second after the cables have separated, the attachment between Dawn and its rocket is severed, and springs push them apart.

http://dawn.jpl.nasa.gov/mission/journal_9_12_07.asp

[pequaide]

Yes I know they work; I have built dozens. But the long tethers working guarantees that r is not a factor. Only mv is conserved, not r * mv. Check their numbers for working tether lengths, and then calculate the velocity needed for r to be a factor. It is impossibly slow.