Wubbly, the 1.100mm should have been written as 1,100mm (1.1m as pequaide corectly understood).
Pequaide, I have done this experiment not only with an open mind but secretly hoping that it will confirm results reported by you, and as result enable us all to seriously thing about ways how to exploit it.
Regrettably the experiment did not confirm what I was hoping for.
I have first clocked the times for the full drop of the weight (1.1m). During this the sound of the dowels moving through air in high speed made me realize that at higher speed of rotation the air friction comes into the play more significantly.
That made me decided to take also the times for the first 3 rotations only.
Angular acceleration will not be constant for the full drop of weight because the air friction decreases the angular acceleration progressively more as the speed increases.
But the times clocked for the first 3 turns (before the speed gets too high) should be resonably reperesentative for evaluating if mr or mrr is corect.
energy producing experiments
Moderator: scott
re: energy producing experiments
I have noted the air resistance. Your speeds are not so high that you will hit a brick wall. And the air resistance will be greater with the mass on the end. The mass on the end is whipping the air and the mass at half radius is moving half as fast for much less air resistance. It is a factor but it will not cause the difference between 3.67 and 5.2. I have used disks.
Atwood's have been known to produce momentum for centuries. Claiming that they produce momentum is within classic physics.
Atwood's have been known to produce momentum for centuries. Claiming that they produce momentum is within classic physics.
re: energy producing experiments
The video is slowed down twice; once when I fed it into the recorder and once more when I recorded it off of the monitor.
There are two video frames where the black squares retain their same angular position.
The cylinder is stopped.
This could be done with a bearing in the center, and with a bearing only one missile needs to be used.
The missile could be released when the cylinder is stopped. Trebuchets do this all the time, and I have done it many times.
Once released the missile has linear motion and this motion is equal to what was the displacement around the arc.
Once released the missile can strike the cylinder in a tangent line. This tangent striking is an experiment that has been done thousands of times. These ballistic pendulum experiments always demonstrate the conservation of linear Newton momentum.
In this experiment 62% of the momentum would be lost if the missiles conserved energy as it swung out and was released from the cylinder.
But energy is not conserved in a tangent striking. And the motion would not be reversible. If energy was first conserved then the motion would be lost and the motion could not be returned to the cylinder.
But apparent in the experiment is that the motion is reversible.
Conclusion; then momentum must be conserved as the missile swings out.
There are two video frames where the black squares retain their same angular position.
The cylinder is stopped.
This could be done with a bearing in the center, and with a bearing only one missile needs to be used.
The missile could be released when the cylinder is stopped. Trebuchets do this all the time, and I have done it many times.
Once released the missile has linear motion and this motion is equal to what was the displacement around the arc.
Once released the missile can strike the cylinder in a tangent line. This tangent striking is an experiment that has been done thousands of times. These ballistic pendulum experiments always demonstrate the conservation of linear Newton momentum.
In this experiment 62% of the momentum would be lost if the missiles conserved energy as it swung out and was released from the cylinder.
But energy is not conserved in a tangent striking. And the motion would not be reversible. If energy was first conserved then the motion would be lost and the motion could not be returned to the cylinder.
But apparent in the experiment is that the motion is reversible.
Conclusion; then momentum must be conserved as the missile swings out.
re: energy producing experiments
92 pages and we're back to where it started. A benchmark of some sort.
re: energy producing experiments
This goes with a video you had been asking for for 92 pages. A frame by frame slow motion video no less. Besides it is a different mass, and it was very interesting.
re: energy producing experiments
Magic round-a-bout is only fun the first few dozen times.
re: energy producing experiments
See: Knex trebuchet; and Knex centrifugal trebuchet
re: energy producing experiments
Knex trebuchet ACDC5968 double arm whipper
re: energy producing experiments
In the tower trebuchet note that the throw begins immediately with little or no loose tether. The ball starts at the bottom and by the time is starts down again it is on the end of a tight tether. Note also that the ball appears to stop the descending drive mass; and when it does is the ball going up or down.
The double arm treb makes a nice throw and there is a whole lot of motion left in the trebuchet.
The double arm treb makes a nice throw and there is a whole lot of motion left in the trebuchet.
re: energy producing experiments
Smokin Lamas 1586.84 ft WR
This is a centrifugal human powered trebuchet.
This is a centrifugal human powered trebuchet.
re: energy producing experiments
Try to imagine 'Inertia II' or 'Bad to the Bone' being driven by a man on a bicycle; kind of silly isn't it. Yet Smokin Lamas throws almost as well.
http://www.youtube.com/watch?v=soVnYzPnRSU
Do you hear the whoosh as Lamas throws. It is throwing in a fluid and the resistance is huge, yet it still throws 1586 feet. Smokin Lamas is conserving momentum not energy as it throws.
Imagine the motor of 'Inertia II' hooked up to Smokin Lamas. Anyone doubt that the speed of sound would be surpassed?
http://www.youtube.com/watch?v=soVnYzPnRSU
Do you hear the whoosh as Lamas throws. It is throwing in a fluid and the resistance is huge, yet it still throws 1586 feet. Smokin Lamas is conserving momentum not energy as it throws.
Imagine the motor of 'Inertia II' hooked up to Smokin Lamas. Anyone doubt that the speed of sound would be surpassed?
re: energy producing experiments
I was trying to tune the mass ratio on a wheel trebuchet. I knew that the tether and mass were too long and too high but I had lost my golf ball in the leaves. The BB bag did something I was unfamiliar with; It stalled in mid air about 300° out. The tether had not yet released of course. I was a bit surprised as the bag stalled so I looked back at the wheel. The tether finally released and the wheel was spinning with what appeared to be the original velocity. But the wheel was spinning in the opposite direction.
The bagged had absorbed all the motion; stopped the wheel; and restarted it in the opposite direction. I had seen restarts a lot but what was interesting was how violently the wheel was spinning. It was quite obvious that the bag had returned almost all the motion to the wheel.
A total return of motion to the wheel would be impossible if the BB bag had conserved the energy of the wheel instead of conserving the momentum. Because we know that only the momentum of a small object can be given to the larger object. If the bag had only the energy of the wheel it would have had only 1/6th or 1/7th of the necessary momentum to restore the motion back into the wheel.
The bagged had absorbed all the motion; stopped the wheel; and restarted it in the opposite direction. I had seen restarts a lot but what was interesting was how violently the wheel was spinning. It was quite obvious that the bag had returned almost all the motion to the wheel.
A total return of motion to the wheel would be impossible if the BB bag had conserved the energy of the wheel instead of conserving the momentum. Because we know that only the momentum of a small object can be given to the larger object. If the bag had only the energy of the wheel it would have had only 1/6th or 1/7th of the necessary momentum to restore the motion back into the wheel.
re: energy producing experiments
I acquired this from Jim-mich; I don't think he will mind. It is the RPM needed for the CF to hold a loose mass on the inside of the rim. But it also shows at what minimal RPM the wheel will throw a mass, on the end of a string, that is on the outside of the rim at 12 o'clock . I prefer throwing at 12 o'clock because by the time the missile starts back up it already contains violent motion. By the time gravity begins to subtract motion, instead of adding motion, the time over which the force can act has become very small.
This will be very helpful to me; thanks again Jim. For example: My 11 inch radius wheel will throw at about 1 RPS.
Speed needed for centrifugal force to equal weight...
1 inch radius = 187.594 RPM
2 inch radius = 132.649 RPM
3 inch radius = 108.307 RPM
4 inch radius = 93.797 RPM
5 inch radius = 83.895 RPM
6 inch radius = 76.585 RPM
7 inch radius = 70.904 RPM
8 inch radius = 66.324 RPM
9 inch radius = 62.531 RPM
10 inch radius = 59.322 RPM
11 inch radius = 56.562 RPM
12 inch radius = 54.154 RPM
13 inch radius = 52.029 RPM
14 inch radius = 50.137 RPM
15 inch radius = 48.437 RPM
16 inch radius = 46.898 RPM
17 inch radius = 45.498 RPM
18 inch radius = 44.216 RPM
19 inch radius = 43.037 RPM
20 inch radius = 41.947 RPM
21 inch radius = 40.936 RPM
22 inch radius = 39.995 RPM
23 inch radius = 39.116 RPM
24 inch radius = 38.292 RPM
25 inch radius = 37.519 RPM
26 inch radius = 36.790 RPM
27 inch radius = 36.102 RPM
27.9 inch radius = 25.113 RPM (1st Wheel - 4.65 ft, turned ~60 RPM)
28 inch radius = 35.452 RPM
29 inch radius = 34.835 RPM
30 inch radius = 34.250 RPM
31 inch radius = 33.693 RPM
32 inch radius = 33.162 RPM
33 inch radius = 32.656 RPM
34 inch radius = 32.172 RPM
35 inch radius = 31.709 RPM
36 inch radius = 31.266 RPM
37 inch radius = 30.840 RPM
38 inch radius = 30.432 RPM
39 inch radius = 30.039 RPM
40 inch radius = 29.661 RPM
41 inch radius = 29.297 RPM
42 inch radius = 28.946 RPM
43 inch radius = 28.608 RPM
44 inch radius = 28.281 RPM
45 inch radius = 27.965 RPM
46 inch radius = 27.659 RPM
47 inch radius = 27.363 RPM
48 inch radius = 27.077 RPM
49 inch radius = 26.799 RPM
50 inch radius = 26.530 RPM
51 inch radius = 26.268 RPM
52 inch radius = 26.015 RPM
53 inch radius = 25.768 RPM
54 inch radius = 25.528 RPM
55 inch radius = 25.295 RPM
55.8 inch radius = 25.113 RPM (2nd Wheel - 9.3 ft, turned ~50 RPM)
56 inch radius = 25.068 RPM
57 inch radius = 24.847 RPM
58 inch radius = 24.632 RPM
59 inch radius = 24.423 RPM
60 inch radius = 24.218 RPM
61 inch radius = 24.019 RPM
62 inch radius = 23.824 RPM
63 inch radius = 23.635 RPM
64 inch radius = 23.449 RPM
65 inch radius = 23.268 RPM
66 inch radius = 23.091 RPM
66.9 inch radius = 22.935 RPM (3rd Wheel - 11.15 ft, turned ~40 RPM)
67 inch radius = 22.918 RPM
68 inch radius = 22.749 RPM
69 inch radius = 22.584 RPM
70 inch radius = 22.422 RPM
71 inch radius = 22.263 RPM
72 inch radius = 22.108 RPM (4th Wheel - 12 ft, turned 26 RPM)
Let 12 o’clock be 360°. That makes 10 seconds after 12 o’clock 1°. Sine of 90° from the horizontal (360°) is 1. Sin of 89° from the horizontal (1°) is .9998477. In a one meter radius vertical wheel the wheel must turn an unattached mass at 360° through .1523 vertical mm or .0001523 m before gravity can take it away from the inside of the circle of the wheel.
Gravity will cause something to fall through .000152305 meters in .00557234 seconds (d = ½ a*t*t); which is one degree in .00557 seconds; or one rotation in (360 * .00557234) 2.0060 seconds or 29.9 RPM.
This concurs with Jim-mich's post
This will be very helpful to me; thanks again Jim. For example: My 11 inch radius wheel will throw at about 1 RPS.
Speed needed for centrifugal force to equal weight...
1 inch radius = 187.594 RPM
2 inch radius = 132.649 RPM
3 inch radius = 108.307 RPM
4 inch radius = 93.797 RPM
5 inch radius = 83.895 RPM
6 inch radius = 76.585 RPM
7 inch radius = 70.904 RPM
8 inch radius = 66.324 RPM
9 inch radius = 62.531 RPM
10 inch radius = 59.322 RPM
11 inch radius = 56.562 RPM
12 inch radius = 54.154 RPM
13 inch radius = 52.029 RPM
14 inch radius = 50.137 RPM
15 inch radius = 48.437 RPM
16 inch radius = 46.898 RPM
17 inch radius = 45.498 RPM
18 inch radius = 44.216 RPM
19 inch radius = 43.037 RPM
20 inch radius = 41.947 RPM
21 inch radius = 40.936 RPM
22 inch radius = 39.995 RPM
23 inch radius = 39.116 RPM
24 inch radius = 38.292 RPM
25 inch radius = 37.519 RPM
26 inch radius = 36.790 RPM
27 inch radius = 36.102 RPM
27.9 inch radius = 25.113 RPM (1st Wheel - 4.65 ft, turned ~60 RPM)
28 inch radius = 35.452 RPM
29 inch radius = 34.835 RPM
30 inch radius = 34.250 RPM
31 inch radius = 33.693 RPM
32 inch radius = 33.162 RPM
33 inch radius = 32.656 RPM
34 inch radius = 32.172 RPM
35 inch radius = 31.709 RPM
36 inch radius = 31.266 RPM
37 inch radius = 30.840 RPM
38 inch radius = 30.432 RPM
39 inch radius = 30.039 RPM
40 inch radius = 29.661 RPM
41 inch radius = 29.297 RPM
42 inch radius = 28.946 RPM
43 inch radius = 28.608 RPM
44 inch radius = 28.281 RPM
45 inch radius = 27.965 RPM
46 inch radius = 27.659 RPM
47 inch radius = 27.363 RPM
48 inch radius = 27.077 RPM
49 inch radius = 26.799 RPM
50 inch radius = 26.530 RPM
51 inch radius = 26.268 RPM
52 inch radius = 26.015 RPM
53 inch radius = 25.768 RPM
54 inch radius = 25.528 RPM
55 inch radius = 25.295 RPM
55.8 inch radius = 25.113 RPM (2nd Wheel - 9.3 ft, turned ~50 RPM)
56 inch radius = 25.068 RPM
57 inch radius = 24.847 RPM
58 inch radius = 24.632 RPM
59 inch radius = 24.423 RPM
60 inch radius = 24.218 RPM
61 inch radius = 24.019 RPM
62 inch radius = 23.824 RPM
63 inch radius = 23.635 RPM
64 inch radius = 23.449 RPM
65 inch radius = 23.268 RPM
66 inch radius = 23.091 RPM
66.9 inch radius = 22.935 RPM (3rd Wheel - 11.15 ft, turned ~40 RPM)
67 inch radius = 22.918 RPM
68 inch radius = 22.749 RPM
69 inch radius = 22.584 RPM
70 inch radius = 22.422 RPM
71 inch radius = 22.263 RPM
72 inch radius = 22.108 RPM (4th Wheel - 12 ft, turned 26 RPM)
Let 12 o’clock be 360°. That makes 10 seconds after 12 o’clock 1°. Sine of 90° from the horizontal (360°) is 1. Sin of 89° from the horizontal (1°) is .9998477. In a one meter radius vertical wheel the wheel must turn an unattached mass at 360° through .1523 vertical mm or .0001523 m before gravity can take it away from the inside of the circle of the wheel.
Gravity will cause something to fall through .000152305 meters in .00557234 seconds (d = ½ a*t*t); which is one degree in .00557 seconds; or one rotation in (360 * .00557234) 2.0060 seconds or 29.9 RPM.
This concurs with Jim-mich's post