This has been mentioned before; but with this video depicting deep space maybe it will seem a more realistic argument.
There are two Identical rockets in deep space. They both have a mass of 5000 kilograms. One rocket is moving 1000 m/sec and the other is at rest. They both fire their thrusters for the same period of time and consume the same amount of fuel (energy), they both have an increase in velocity of one meter per second.
The slow rocket has an energy change of 2500 joules; the other has and energy change of 5,002,500 joules. Five million joule difference for the same amount of fuel???
energy producing experiments
Moderator: scott
re: energy producing experiments
http://makezine.com/projects/make-25/th ... -windlass/
I am surprised that people can get something wrong; but then they don’t step back and compel themselves to get it right. I am thinking of this description of a Chinese windlass.
They clearly state that it takes 12.5 pounds to lift 600 pounds (apparently determined by experiment); which is a mechanical advantage of 48. But their formula gives a mechanical advantage of 96. Why don’t they figure out what is wrong. Why do you divide by 2?
One pulley can give you a mechanical advantage of 2 if the ceiling is holding the rope on the other side. Then the puller (force) will lift twice as far as the object moves, but with half the force. But the puller in this situation is lifting both sides of the rope going through the pulley; the one side holds the other. The one side holds while the other side lifts?
They treat the problem as if it were a circumference issue, or used rope length to load lift distance. This is legitimate, but with an application of a little algebra it is easier than that. They use crank radius (cR) to the large axle radius and then use circumference to circumference of the two axle diameters.Â
Well a circumference C / C -c is D * pi / (D - d) * pi; and the pi’s drop out. Now we have D / D-d which is R * 2 /(R - r)2 and the twos drop out, so we have R / R- r. So the mechanical advantage is Crank(cR) / R * R /R-r. Now an R in the numerator and denominator can drop out. That leaves us with crank radius over axle radius difference. Or 6/.125 = 48. Times the lower pulley 2; but why do we then divide by two?
But now; for the main point. This is a dynamic system that is mr not mr². The mr does not just apply to a static system it applies to a dynamic system. The windlass will lift bridges and the formula is crank radius over the functional radius of the axle; R/r. Or mR/Mr or Fr/fR.
If the crank lever were a wheel, then 13 pounds of force applied to the wheel would uniformly accelerate the load mass upward. If that wheel had an application of 12 pounds of force then the load mass would accelerate downward. Both would have a mechanical advantage (or disadvantage) of 48.
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It is dynamic and it uses an mr formula. Not mr².
Either the 12 lbs or the 13 lbs will be moving 48 times faster than the 600 lbs (272 kg). So you have ½ * 600 / 2.207 * 1 * 1 = 136 joules; or ½ * 12 */ (2.207 lb/kg) * 48 * 48 = 6263 J. And remember the center of mass between the 12.5 and the 600 lbs does not change; only the imbalance is dropped.
So another energy producing machine; that the Chinese had, who knows how many centuries ago.
I am surprised that people can get something wrong; but then they don’t step back and compel themselves to get it right. I am thinking of this description of a Chinese windlass.
They clearly state that it takes 12.5 pounds to lift 600 pounds (apparently determined by experiment); which is a mechanical advantage of 48. But their formula gives a mechanical advantage of 96. Why don’t they figure out what is wrong. Why do you divide by 2?
One pulley can give you a mechanical advantage of 2 if the ceiling is holding the rope on the other side. Then the puller (force) will lift twice as far as the object moves, but with half the force. But the puller in this situation is lifting both sides of the rope going through the pulley; the one side holds the other. The one side holds while the other side lifts?
They treat the problem as if it were a circumference issue, or used rope length to load lift distance. This is legitimate, but with an application of a little algebra it is easier than that. They use crank radius (cR) to the large axle radius and then use circumference to circumference of the two axle diameters.Â
Well a circumference C / C -c is D * pi / (D - d) * pi; and the pi’s drop out. Now we have D / D-d which is R * 2 /(R - r)2 and the twos drop out, so we have R / R- r. So the mechanical advantage is Crank(cR) / R * R /R-r. Now an R in the numerator and denominator can drop out. That leaves us with crank radius over axle radius difference. Or 6/.125 = 48. Times the lower pulley 2; but why do we then divide by two?
But now; for the main point. This is a dynamic system that is mr not mr². The mr does not just apply to a static system it applies to a dynamic system. The windlass will lift bridges and the formula is crank radius over the functional radius of the axle; R/r. Or mR/Mr or Fr/fR.
If the crank lever were a wheel, then 13 pounds of force applied to the wheel would uniformly accelerate the load mass upward. If that wheel had an application of 12 pounds of force then the load mass would accelerate downward. Both would have a mechanical advantage (or disadvantage) of 48.
Â
It is dynamic and it uses an mr formula. Not mr².
Either the 12 lbs or the 13 lbs will be moving 48 times faster than the 600 lbs (272 kg). So you have ½ * 600 / 2.207 * 1 * 1 = 136 joules; or ½ * 12 */ (2.207 lb/kg) * 48 * 48 = 6263 J. And remember the center of mass between the 12.5 and the 600 lbs does not change; only the imbalance is dropped.
So another energy producing machine; that the Chinese had, who knows how many centuries ago.
re: energy producing experiments
Peq.
Once again you are mixing up inertia and torque. Torque relies on leverage ration, or MR in certain circumstances. Whenever you see the phrase, "mechanical advantage", it refers to torque. The windlass didn't create energy, it transformed it in a way that conserves energy: input force x distance = output force x distance.
You are welcome.
K.
Once again you are mixing up inertia and torque. Torque relies on leverage ration, or MR in certain circumstances. Whenever you see the phrase, "mechanical advantage", it refers to torque. The windlass didn't create energy, it transformed it in a way that conserves energy: input force x distance = output force x distance.
You are welcome.
K.
re: energy producing experiments
T please do not post on this thread, thank you.
Those who fabricate data are not welcome.
Those who fabricate data are not welcome.
re: energy producing experiments
The Chinese no doubt counter balanced their bridges and the windlass was used to overcome an imbalance that was intentionally left in the bridge so that it would close. Lets use 300 kg imbalance and a 50/1 mechanical advantage. That means the peasant is cranking 6 kilograms (58.86 N). The peasant would not appreciate the energy producing capabilities of the system unless he lost control of the crank handle. The hand would soon be moving a lethal speeds.
Let say the bridge was moving one meter per second as it slammed back into position. Then the one kilograms crank would be moving 50 m/sec. And would have 1250 joules of energy. The imbalance in the bridge itself would only have had 150 joules.
The Chinese were interested in lifting bridges and had no curiosity in the energy of the crank. But the energy is there to be used.
Let say the bridge was moving one meter per second as it slammed back into position. Then the one kilograms crank would be moving 50 m/sec. And would have 1250 joules of energy. The imbalance in the bridge itself would only have had 150 joules.
The Chinese were interested in lifting bridges and had no curiosity in the energy of the crank. But the energy is there to be used.
re: energy producing experiments
I constructed a differential windlass out of PVC pipe to get a better idea how they work and to get a 'hands on' understanding. I used a pipe and a T. The pipe had an outside diameter of 1.33 inch (33.6 mm) and the T have a 1.625 (41 mm) inch outside diameter. This gives us (41 * pi – 33.6 * pi) = 23.25 mm per turn. The string is shortened by 23.25 mm per turn; but the string goes down and comes back up so the mass is only lifted 11.62 mm.
I turned the windlass 15 times and the mass was brought up 179.5 mm, Experimentally it should have been 11.62 * 15 =174.4 mm. Close enough.
Fifteen turns for the outside diameter of the T connector would be 41mm * 15 * pi = 1932.1 mm. 1932 /174.4 is a mechanical advantage of 11.08. If the t was connected to a 4 *1.62 = 6.48 inch crank handle you would have a mechanical advantage of 44.3.
This is interesting because the 6 inch crank on the 3 inch over 2.75 inch shafts was roughly the same. And it should be because it is the difference (3 – 2.75) in the shaft diameters not the overall diameter. The 3 inch over 2.75 inch has a difference of .25 inches and the 1.62 inch over 1.33 inch has a difference of .29. their respective (4x) crank handles are 6 and 6.48. And their mechanical advantages are 48 and 44.3.
This seems like a very good way to build speed. Instead of having the large mass on a small shaft (.25 or .29) you can suspend it from a much larger shaft (2.75 or 1.62).
I turned the windlass 15 times and the mass was brought up 179.5 mm, Experimentally it should have been 11.62 * 15 =174.4 mm. Close enough.
Fifteen turns for the outside diameter of the T connector would be 41mm * 15 * pi = 1932.1 mm. 1932 /174.4 is a mechanical advantage of 11.08. If the t was connected to a 4 *1.62 = 6.48 inch crank handle you would have a mechanical advantage of 44.3.
This is interesting because the 6 inch crank on the 3 inch over 2.75 inch shafts was roughly the same. And it should be because it is the difference (3 – 2.75) in the shaft diameters not the overall diameter. The 3 inch over 2.75 inch has a difference of .25 inches and the 1.62 inch over 1.33 inch has a difference of .29. their respective (4x) crank handles are 6 and 6.48. And their mechanical advantages are 48 and 44.3.
This seems like a very good way to build speed. Instead of having the large mass on a small shaft (.25 or .29) you can suspend it from a much larger shaft (2.75 or 1.62).
re: energy producing experiments
I was looking up ‘moment of inertia’ to figure out how such a ridicules concept made it into the world of physics. I saw that ‘moment of inertia’ is attributed to a man named Euler. It appears that Euler was a prime example of an academic; obviously he never submitted his ‘moment of inertia’ theory to experimentation.
The origin of the name Euler also perked my interest; is that a German name? Because I remember another German that ripped Newton; all but calling him an idiot, ‘Leibnitz’ despised Newton. I found that Euler was Swiss and studied in Berlin.
Is it possible that the political structure of Continental Europe was interested in discrediting all things Newtonian (British)?
The time period is interesting too; the middle 18th century is when Newtonian Physics was supplanted by Leibnitzian Physics. Liebnitz’s mv² is a direct violation of Newtonian physics and Leibniz delivered written animosity toward Newton for his mv. Was Euler’s mr² also used to discredit Newton?
Euler’s ‘moment of inertia’ states that a mass on the end of a lever with 10 units of radius is 100 times harder to rotate than the same mass at one unit of radius. Laws of levers disagrees; and experiments would have proven that this is a false statement. Is it possible that the political structure of the day enhanced ‘moment of inertia’s acceptance?
And ‘moment of inertia’ is part and parcel to another false concept ‘Angular momentum - conservation’ (in the lab), which is also not Newtonian.
No comments are necessary; unless you can find Liebnitz's tirade against Newton, I would not mind reading that again, good for a few laughs.
The origin of the name Euler also perked my interest; is that a German name? Because I remember another German that ripped Newton; all but calling him an idiot, ‘Leibnitz’ despised Newton. I found that Euler was Swiss and studied in Berlin.
Is it possible that the political structure of Continental Europe was interested in discrediting all things Newtonian (British)?
The time period is interesting too; the middle 18th century is when Newtonian Physics was supplanted by Leibnitzian Physics. Liebnitz’s mv² is a direct violation of Newtonian physics and Leibniz delivered written animosity toward Newton for his mv. Was Euler’s mr² also used to discredit Newton?
Euler’s ‘moment of inertia’ states that a mass on the end of a lever with 10 units of radius is 100 times harder to rotate than the same mass at one unit of radius. Laws of levers disagrees; and experiments would have proven that this is a false statement. Is it possible that the political structure of the day enhanced ‘moment of inertia’s acceptance?
And ‘moment of inertia’ is part and parcel to another false concept ‘Angular momentum - conservation’ (in the lab), which is also not Newtonian.
No comments are necessary; unless you can find Liebnitz's tirade against Newton, I would not mind reading that again, good for a few laughs.
re: energy producing experiments
Draw a circle on a table top. Place a toy truck or car on the circle. Place a pin in the middle of the circle and use that point as the fulcrum; or point of rotation. Tape a dowel end to the truck. Contact the other end of the dowel with the pin. Rotate the truck on the circle by placing your finger one tenth of the distance from the pivot point to the truck.
The distance from the finger to the pin is the effort arm.
The distance from the pin to the truck is the resistance arm.
In the mr² formula the r would be 10 and it would be 100 times as difficult to get the same rate of rotation as if the truck were at the position of the finger. The finger would have to push 100 times harder to get the same rotation. Laws of levers says 10 times harder. If you have an air table try the same experiment with a puck.
Now lets replace the dowel with a string.
Now lets place the same force one tenth the distance down the string.
And: Oh: wait a minute. The truck no longer moves in the circle. The truck is pulled toward the point of rotation. We have no effort arm and no resistance arm along the length of the string as previously described. The truck is moving slower than the force, this has no relationship to arm lengths and circular motion. The academicians apply mr² to strings but it is a totally unrelated composition of objects. The cable attached to a winch is not a radius. And the cable attached to the mass on the Yo-yo despin device is not a radius either.
The radius for the cable in the yo-yo despin device is the radius of the unwind cylinder. The cable itself can not be used to produce a torque on the center of motion; its working radius is only the radius of the unwind cylinder. The resistance arm in the yo-yo despin device is almost equal to the radius of the unwind cylinder. The r's would be roughly equal; or one to one.
The mass on the end of the cable does not have massive amounts of torque: its working radius is no larger than the resistance radius. So you have a false formula (mr²) under false conditions.
The distance from the finger to the pin is the effort arm.
The distance from the pin to the truck is the resistance arm.
In the mr² formula the r would be 10 and it would be 100 times as difficult to get the same rate of rotation as if the truck were at the position of the finger. The finger would have to push 100 times harder to get the same rotation. Laws of levers says 10 times harder. If you have an air table try the same experiment with a puck.
Now lets replace the dowel with a string.
Now lets place the same force one tenth the distance down the string.
And: Oh: wait a minute. The truck no longer moves in the circle. The truck is pulled toward the point of rotation. We have no effort arm and no resistance arm along the length of the string as previously described. The truck is moving slower than the force, this has no relationship to arm lengths and circular motion. The academicians apply mr² to strings but it is a totally unrelated composition of objects. The cable attached to a winch is not a radius. And the cable attached to the mass on the Yo-yo despin device is not a radius either.
The radius for the cable in the yo-yo despin device is the radius of the unwind cylinder. The cable itself can not be used to produce a torque on the center of motion; its working radius is only the radius of the unwind cylinder. The resistance arm in the yo-yo despin device is almost equal to the radius of the unwind cylinder. The r's would be roughly equal; or one to one.
The mass on the end of the cable does not have massive amounts of torque: its working radius is no larger than the resistance radius. So you have a false formula (mr²) under false conditions.
re: energy producing experiments
I drilled a hole in the center of a one inch square HDPE bar 34 inches long. With 17 inches on each side this allowed a teeter totter arrangement for the drill bit placed through the hole.
I placed a compression force gauge at about 11 inches on the top of the one side. On the other side I placed 902 grams at about 1.6 inches. The force gauge showed 1.0 newtons . I moved the force gauge up and down and the gauge would read 1.0 and 1.5 N. It took force to make the 902 grams go up, but only about .5 N.
In the same arrangement I had a compression force gauge at about 11 inches on one side. On the other side I placed 90 grams at about 16 inches. The force gauge showed 1.0 newtons . I moved the force gauge up and down and the gauge would read 1.0 and 1.5 N. It took force to make the 90 grams go up, but only about .5 N.
Conclusion: it takes no more force to make 90 grams at 16 inches move than it takes to make 902 grams at 1.6 inches move. As Euclid would say: 900 grams at 1.6 inches is equal to 90 grams at 16 inches because they are both equal to the same reading of 1.5 newtons on the same force gauge with the same distance from the fulcrum.
Then I placed the 90 grams on the end of a 30 inch string and suspended it from the same 16 inch location. The gauge did not show massive amounts of angular momentum as the 90 grams swung back and forth while I moved the gauge up and down. The gauge keep reading 1.0 to 1.5 N.
The oppositional theory of Euler mr² would have given 15 N on the gauge for the 90 grams at 16 inches. But he could not support his theory by conducting experiment; and no one bothered to challenge him.
I placed a compression force gauge at about 11 inches on the top of the one side. On the other side I placed 902 grams at about 1.6 inches. The force gauge showed 1.0 newtons . I moved the force gauge up and down and the gauge would read 1.0 and 1.5 N. It took force to make the 902 grams go up, but only about .5 N.
In the same arrangement I had a compression force gauge at about 11 inches on one side. On the other side I placed 90 grams at about 16 inches. The force gauge showed 1.0 newtons . I moved the force gauge up and down and the gauge would read 1.0 and 1.5 N. It took force to make the 90 grams go up, but only about .5 N.
Conclusion: it takes no more force to make 90 grams at 16 inches move than it takes to make 902 grams at 1.6 inches move. As Euclid would say: 900 grams at 1.6 inches is equal to 90 grams at 16 inches because they are both equal to the same reading of 1.5 newtons on the same force gauge with the same distance from the fulcrum.
Then I placed the 90 grams on the end of a 30 inch string and suspended it from the same 16 inch location. The gauge did not show massive amounts of angular momentum as the 90 grams swung back and forth while I moved the gauge up and down. The gauge keep reading 1.0 to 1.5 N.
The oppositional theory of Euler mr² would have given 15 N on the gauge for the 90 grams at 16 inches. But he could not support his theory by conducting experiment; and no one bothered to challenge him.
re: energy producing experiments
Just for confirmation I placed 130 grams at 11 inches on the above experiment. The force gauge stayed at the 1.0 N reading. The effort arm is also at 11 inches. The smallest increment of measurement for the gauge is .5 N.
So 11in * 130g; 16in * 90g and 1.6in * 902g are all rotated at the same rate; when the same effort force is placed at the same 11 inch radius.
There is no mechanical advantage when the effort arm and the resistance arm have the same radius; so this is where you get a direct acceleration from the F = ma relationship. You can directly plug in your force and mass numbers to get acceleration. So with a certain force you get a certain acceleration; and because the motion is captured in a circle a certain rotation. The force causes each mass to comply with the F = ma formula; no matter what the radial position of the mass.
Whatever the rate; the force cannot tell the difference between 130 grams at 11 inches, or 90 grams at 16 inches, or 902 grams at 1.6 inches: they all rotate at the same rate. At one meter per second around the circle, at the applied force, 130 grams will have .130 units of momentum. The 90 gram will be moving 16/11 faster at 1.45 m/sec and it will have .131 units of momentum. The 902 grams will be going 1.6/11 slower at .145 m/sec for .131 units of momentum. The 902 grams will have .00951 joules of energy. The 130 grams will have .065 joules. And the 90 grams will have .0946 joules.
Now let’s apply this information to another experiment (Bob Stein rotary trebuchet). A uniform quantity of force is placed at a radius; and that force will give the wheel an F = ma acceleration. At all points it is F = ma; at all point outside the radius, as is the 90 grams, it is F =ma. That all points inside the radius, as is the 902 grams, it is F = ma. All points of mass on the circle of the plywood disk are given momentum: not energy.
When the missile is released a reverse force is applied at the circumference of the disk; but the same rules apply. All points inside the radius have momentum removed by the applied force (F = ma). With Newton Third Laws in play the missile must contain the momentum that the disk has lost.
I also balanced the bar and placed two extra grams on one end; the two grams accelerated a 90 gram mass at 16 inch just as easily as a 900 grams mass at 1.6 inches. At 11 inches this would be a simple Atwood's; but it is also F = ma at the other radii. They all have the same momentum.
So 11in * 130g; 16in * 90g and 1.6in * 902g are all rotated at the same rate; when the same effort force is placed at the same 11 inch radius.
There is no mechanical advantage when the effort arm and the resistance arm have the same radius; so this is where you get a direct acceleration from the F = ma relationship. You can directly plug in your force and mass numbers to get acceleration. So with a certain force you get a certain acceleration; and because the motion is captured in a circle a certain rotation. The force causes each mass to comply with the F = ma formula; no matter what the radial position of the mass.
Whatever the rate; the force cannot tell the difference between 130 grams at 11 inches, or 90 grams at 16 inches, or 902 grams at 1.6 inches: they all rotate at the same rate. At one meter per second around the circle, at the applied force, 130 grams will have .130 units of momentum. The 90 gram will be moving 16/11 faster at 1.45 m/sec and it will have .131 units of momentum. The 902 grams will be going 1.6/11 slower at .145 m/sec for .131 units of momentum. The 902 grams will have .00951 joules of energy. The 130 grams will have .065 joules. And the 90 grams will have .0946 joules.
Now let’s apply this information to another experiment (Bob Stein rotary trebuchet). A uniform quantity of force is placed at a radius; and that force will give the wheel an F = ma acceleration. At all points it is F = ma; at all point outside the radius, as is the 90 grams, it is F =ma. That all points inside the radius, as is the 902 grams, it is F = ma. All points of mass on the circle of the plywood disk are given momentum: not energy.
When the missile is released a reverse force is applied at the circumference of the disk; but the same rules apply. All points inside the radius have momentum removed by the applied force (F = ma). With Newton Third Laws in play the missile must contain the momentum that the disk has lost.
I also balanced the bar and placed two extra grams on one end; the two grams accelerated a 90 gram mass at 16 inch just as easily as a 900 grams mass at 1.6 inches. At 11 inches this would be a simple Atwood's; but it is also F = ma at the other radii. They all have the same momentum.
re: energy producing experiments
The cylinder and sphere I am working with now is a pipe with a hole through a diameter. A one inch sphere is seated in each hole. They are connected by a string that is about 8.28 r long.
In operation the spheres jump away from the cylinder and stop it in about one third rotation, the spheres appear to be rotating about 4 times as fast as the original speed of the cylinder and spheres before release. They are on a string 4 times as long for a linear v of about 16. This is consistent with the spheres mass to 'cylinder and spheres' mass ratio of about 1 to 15.5.
I have made arrangements to use a corporation's high speed high def. camera. I wonder if they can see it and not see it.
In operation the spheres jump away from the cylinder and stop it in about one third rotation, the spheres appear to be rotating about 4 times as fast as the original speed of the cylinder and spheres before release. They are on a string 4 times as long for a linear v of about 16. This is consistent with the spheres mass to 'cylinder and spheres' mass ratio of about 1 to 15.5.
I have made arrangements to use a corporation's high speed high def. camera. I wonder if they can see it and not see it.
re: energy producing experiments
Yes; I think so too. I was sorry to see pumpkin chunkin got canceled this year. I was looking forward to seeing what progress they had made in the centrifugal s.