Idea I had since elementary school

[Tarsier79]

Firstly, your L lever:

Unfortunately, I have a new computer, and my version of photoshop is not compatible with win7. So MS paint here we go...

Assuming the levers are weightless, the COM will lie somewhere on the green line. the brown weight is the approximate position of the COM if your weight ratio is 1:10.( To find the position, divide that line by 11, and your first notch will be the COM) The COM will be fall to directly below the pivot, so we can change the parameters for the L lever to lift above by dropping the ramp, or enlarging the entire L lever mech.

To calculate the gear ratio, you just need to know the angles the two levers move through, from measurements taken from this somewhat blurry animation, you can measure/calc your lever movement angles are approximately 90 and 16 degrees so your gear ratio is approx 1 : 5.625 .
Then If you calculate your lever ratio, it needs to be divided by the gear ratio.


Ill have to leave it there for the moment, but will show you the extended lever calcs later if someone doesn't beat me to the punch.

[preoccupied]

What is the gear ratio if the ramp was perfectly level? Well the right angle bar turns 135 degrees because the dimensions for radiusshift11 animation is length of the right angle bar 2inchesx2inches and the length up to the long lever 2 inches. It's driven/driver so for the long lever pulling down using the weight it would be long lever's degree/135 degrees. If the ramp was perfectly level and the long lever is 20 inches the long lever would turn 11.53 degrees. 11.53/135=.08 - that's discouraging. I'm glad though that I have figured out how to calculate the gear ratio using that online trig calculator James shared in the post. If the long lever was 100 inches the ratio would be 2.29/135=.016. F2*D2=F1*D1 so for .08 gear ratio, 1N*.08*20inches=1.6 and 2*2.83=5.66. The 1.6 is less than the 5.66 so it won't budge at 20 inches. For 200 inches 1N*.016*100=1.6 and 2*2.83=5.66 and again is the same. Funny1.gif therefore will not budge no matter how long the lever is because the gear ratio will always be too high.

EDIT
I did the math wrong but the general conclusion should be the same. The long lever will budge but it will not go all the way no matter how long the ramp is. A longer ramp might not be a waste because a slightly nearer result might be achieved but not quite all the way. I don't know because I can't calculate the torque unless it is perfectly level because I learned how to calculate torque on a simple lever from a youtube video.

[Tarsier79]

Here is the relationship between how long you make the lever, and its power.


In the three levers below, notice all 3 are allowed to only drop a distance of 1 unit.

The top lever has a ratio of 1:1, so the blue 1kg weight is balanced by a 1kg weight moving a distance of 1 also.

The middle lever is doubled in length, so the green 1kg weight is balanced by a 2kg weight, which moves a distance of .5

The lower lever is 4x the top, so the red weight is balanced by a 4kg weight moving a distance of .25 units.


So, no matter how much you extend the lever, its lifting power is the same. The weight x Height of all 3 levers = 1kg x 1 unit. Regardless of the lever length the common factor is the drive weight (1kg) drops 1 unit. Because gravity acts vertically only, the only consideration is the drive weight x vertical movement vs load weight x vertical movement. Do you see how this relates to your mechanism?

If your large weight is capable of lifting your small weight a distance of X, then it rolls down a ramp to where its height is less, how do you expect to increase its power enough to reset the larger weight?....




EDIT: I am glad you came to the same conclusion as I was posting my explanation. You are right, the mechanism will not reset.

[Tarsier79]

Would you like to "uncomplicate" things?

You stated yourself that the length of the long lever is irrelevant. What if the angle the "long" lever rotates is the same as the angle the large weight drives the smaller weight?.... You would have a gear ratio of 1:1.

Now what do you think would happen?

[preoccupied]

Tarsier79, I had to edit my post because I was wrong. The long lever and the gear ratio are relevant but no matter how long the long lever becomes it will not reach the destination at the end of its swing. Or I could be wrong in my logic now but anyways I did the calculations wrong in the last post. These new calculations for radiusshift11 should be correct.

I tried to calculate where the radiusshift11 animation would come into balance during the long lever's swing. I looked at wikki and it said perpendicular to the fulcrum for M=F*d. Using this I can now calculate the multiple radiusshift11's connected to each other. First I have to calculate where the first machine comes into balance. I will have to use guess and check. If the long lever travels 45 degrees then the short lever would travel 10/7 or 64.29 degrees. The distance perpendicular to the fulcrum for 45 degrees using the long lever is 3.16 so F2*D2=F1*D1 is 1N*3.16*7/10=2.212. For 64.29 degrees and 2.83 length of the L shaped lever on the heavy weight's side would have the heavy weight be a distance perpendicalar to the fulcrum of 2.54. F2*D2=F1*D1 is 2N*2.54=5.08. 45 degrees is too far for the long lever to go. 30 degrees on the long lever would have moved the L shaped lever 10/7 gear ratio or 42.85. The distance for the long lever from the fulcrum is 3.87 and the L shaped lever from the fulcrum is 1.92. For F2*D2=F1*D1, 3.87*(7/10)*1N=2.709 and 1.92*2=3.84. 30 degrees for the long lever is still too far. Lets try 10 degrees for the long lever and 10/7 or 14.28. The distance for the long lever from the fulcrum is 4.4 and the distance for the L shaped lever from the fulcrum is .69. .69D*2N=1.38N and 4.4d*1N*.7=3.08 so 10 degrees is too short of a fall to balance it. Lets try 20 degrees for the long lever and 10/7 or 28.57 degrees for the L shaped lever. For the L shaped lever the distance is 1.35 and for the long lever the distance is 4.2. For F2*D2=F1*D1, 1N*.7*4.2=2.94 and 1.35*2=2.7. The long lever is too high. Lets try 25 degrees for the long lever and 10/7 or 35.71. For the L shaped lever the distance is 1.65 and the long lever distance is 2.56. 2.56*1N*.7=1.79 and 1.65*2=3.3. The tables have turned and fast! Lets try 23.5 degrees on the long lever and 10/7 or 33.57. The distance for the long lever is 4.1 and the distance for the L shaped lever is1.56. 4.1*1N*.7=2.87 and 1.56*2=3.12. For 23 degrees the long lever is 4.11 distance and the L shaped lever is 32.85 degrees and 1.53 distance. 4.11*.7=2.87 and 2*1.53=3.06. 22 degrees for the long lever is 4.14 and 31.43 degrees and 1.47 distance for the L shaped lever. 1N*.7*4.14=2.9 and 1.47*2=2.94. Very close we are almost there! 21.8 degrees is 4.15 distance with the long lever and 31.14 degrees and 1.46 distance for the L shaped lever. 1N*.7*4.15=2.9 and 2*1.46=2.92. That's close enough. The long lever can go about 21.8 degrees before balancing out. But just to make sure I will do 21.7. 21.7 is 4.15 distance and 31 is 1.45 distance. 1N*.7*4.15=2.9 and 1.45*2=2.9. Wow I almost got ahead of myself with the 21.8 conclusion because 21.7 fits perfectly! The weights balance at 21.7 degrees. Now I need to figure out how the weight balance when there are two machines connected to each other.

[preoccupied]

Torque on first machine after coming into balance and the new machine initially adds a weight ot its long lever .7*4.15+.5(.7*4.47)-.5(.7*4.15)=4.58 and 1.45*2/2+0=.725
Torque on the second machine with the newly utilized long lever .5(.7*4.47)+.5(.7*4.15)=4.58 and 0+1.45/2=.725.... From this point we can see that there will be movement but where will it come into balance? The long levers will both move the same distance and the L shaped levers will both move 10/7.

Lets see where they would be if the second Machine's long lever is at 20 degrees. The first machine's long lever would be at 41.7 degrees. The second machine's L shaped lever would be at 10/7*20=28.57 degrees and the L shaped lever for the first machine would be at 10/7*41.7=59.57 degrees. .5(.7*3.33)+.5(.7*4.2)=2.64 and .5(2.44*2)+.5(1.35*2)=3.79 so 20 degrees for the new long lever is too far. Lets try 15 degrees, then 15 degrees on the long lever would be 26.7 degrees on the other long lever and the first L shaped lever would be 26.7*(10/7)=38.14 and second L shaped lever would be 15(10/7)=21.43. .5(.7*4.31)+.5(3.99*.7)=2.9 and .5(1.75*2)+.5(1.03*2)=2.78 so the long lever needs to drop more. Try 16 degrees, then the first long lever would be 27.7 and the first L shaped lever would be 27.7(10/7)=39.57 and second L shaped lever would be 16(10/7)=22.85. .5(1.8*2)+.5(1.09*2)=2.96 and .5(.7*4.29)+.5(.7*3.95)=2.88. That's too far so lets try 15.5 degrees, then the first long lever would be 27.3 degrees and the first L shaped lever would be 27.3(10/7)=39 and the second L shaped lever would be 15.5(10/7)=22.14. .5(.7*4.3)+.5(.7*3.97)=2.89 and .5(2*1.78)+.5(2*1.06)=2.84... that's it! that's as close as I need to know. That is where it comes into balance with two machines attached together. The first long lever has reached 27.3 degrees and the first long lever has reached 15.5 degrees. The first long lever must reach 63 degrees to complete the task. We are a little under half way there. Does that mean that a little over twice the machines can complete the task? Tune in for the next post for the thrilling conclusion!

[preoccupied]

The first long lever has reached 27.3 degrees and the second long lever has reached 15.5 degrees. The third long lever starts at 0 degrees and has a length of 4.47. If the third long lever is at 10 degrees then 14.29 37.3 53.29 25.5 36.43 (.7*4.4)/3+(.7*3.55)/3+(.7*4.03)/3=1.03+.83+.94=2.8 and (2*.7)/3+(2*2.26)/3+(1.68*2)/3=.47+1.5+1.12=3.09. The long lever has gone too far. The third long lever is at 8 degrees then 11.42 35.3 50.42 23.5 33.57 (.7*7.39)/3+(.7*5.94)/3+(.7*6.85)/3=1.72+1.39+1.6=4.71 and (2*.56)/3+(2*2.18)/3+(2*1.56)/3=.37+1.45+1.04=2.86. The long lever has not gone far enough. The long lever is 9 degrees then 12.85 34.3 49 22.5 32.14 (.7*4.41)/3+(.7*3.69)/3+(.7*4.12)/3=1.03+.96+.86=2.85 and (2*.62)/3+(2*2.13)/3+(2*1.5)/3=.41+1.42+1=2.83. That's close enough. The third long lever is 9 degrees the second long lever is 22.5 degrees and the first long lever is 34.3 degrees. The first long lever must be 63 degrees.

EDIT

The fourth long lever starts at 0 degrees and has a length of 4.47. If the fourth long lever is 9 degrees then 12.85 18 25.71 31.5 45 43.3 61.85 (.7*4.41)/4+(.7*4.25)/4+(.7*3.81)/4+(.7*3.25)/4=.77+.74+.66+.57=2.74 and (2*.62)/4+(2*1.22)/4+(2*2)/4+(2*2.49)/4=.31+.61+1+1.25=3.17. The long lever has gone too far. If the fourth long lever is 8 degrees then 11.42 17 24.28 30.5 43.57 42.3 60.42 (.7*4.43)/4+(.7*4.27)/4+(.7*3.85)/4+(.7*3.3)/4=.78+.74+.67+.58=2.77 and (2*.56)/4+(2*1.16)+(2*1.95)/4+(2*2.46)/4=.28+.58+.97+1.23=3.06. The fourth long lever has gone too far. If the fourth long lever is 7 degrees then 10 16 22.85 29.5 42.14 41.3 59 (.7*4.43)/4+(.7*4.3)/4+(.7*3.89)/4+(.7*2.95)/4=.78+.75+.68+.52=2.73 and (2*.49)/4+(2*1.09)/4)+(2*1.89)/4+(2*2.42)/4=.25+.55+.94+1.21=2.95. 7 degrees is still too far. Lets try 6 degrees then 8.57 15 21.43 28.5 40.71 40.3 57.57 (.7*4.45)/4+(.7*4.31)/4+(.7*3.93)+(.7*3.4)/4=.78+.75+.69+.6=2.82 and (2*.42)/4+(2*1.03)/4+(2*1.85)/4+(2*2.39)/4=.21+.52+.92+1.19=2.84. That's close enough. The fourth lever reaches about 7 degrees. The third long lever 15 degrees, the second 28.5 and the first long lever reaches 40.3 degrees. The first long lever must reach 63 degrees to complete the task.
More to come!

[preoccupied]

The strongest part of the long lever is (.7*4.47)=3.129 and the strongest angle of the L shaped lever is (2*2.83)=5.66. So with radiusshift11 animation it is impossible to complete the task by adding additional machines using weights 1 and 2 pounds. Maybe if it were a different weight combination it could be more effective. (1.5*.7*4.47)=4.69 no so maybe (1.8*.7*4.47)=5.63 no so maybe (1.9*.7*4.47)=5.94. yes so maybe if the weights can be 1.9 and 2 then it could work with multiple machines. But would the L shaped lever lift a 1.9 weight? The strongest L shaped side would be 1.9*2.83=5.37 and 2*2=4 so no 1.5*2.83=4.245 still too heavy 1.4*2.83=3.962 so just barely be able to lift the weight with 1.4 but 1.4 is not enough to cause the long lever to be able to complete the task with multiple machines. Big fail on radiusshift11. It won't work as designed in the hypothesis. But I have thought of a new way to improve on it. I can make the gear ratio equal and add distance to the long lever at the same time! What is this sorcery you ask? Tune in for the next post to find out!

[preoccupied]

Lets make the big weight 2Pounds and the small weight 1.4 pounds because that is most efficient. We start at the beginning at 3 degrees out so (.15*2)=.3 and (.21*1.4)=.94. At the beginning it doesn't seem to want to move. Maybe a little bit further at 10 degrees out so .69*1.4=.97 and .49*2=.98. Ahh so I'm at a road block. It is about equal.

[preoccupied]

The long lever's distance would be 4.62 and the gear ratio would be 7/10 because the long lever starts at 30 degrees. Instead of the strongest position being at the beginning the strongest position is at the bottom for the long lever. At 30 degrees the L shaped lever would be at about 3 degrees so (1.4*7*2.31)=2.26 and (2*.15)=.3. The long lever starts to move. At the end of the swing (1.4*.7*4.62)=4.52 and (2*2.83)=5.66. So where does it come into balance? I'm going to add a second machine. 2.26+4.52=6.78>5.66 so it might work! Check 50 degrees on the long lever so (3.54*1.4*.7)=3.47 and (2*1.53)=3.06. The long lever has not gone far enough. Check 58 degrees so (2*1.97)=3.94 and (.7*1.4*3.91)=3.83. Close but no cigar. Check 57 degrees (.7*1.4*3.87)=3.79 and (2*1.92)=3.84. Still no cigar. Check 56 degrees (.7*1.4*3.83)=3.75 and (2*1.87)=3.74. That's where it comes into balance 56 degrees on the long lever.

Check 47 degrees so (.98*3.37)/2+(.98*4.42)/2=1.65+2.17=3.82 and (2*1.35)/2+(2*2.57)/2=1.35+2.57=3.92. The second long lever goes too far. Check 46 degrees so (.98*3.32)/2+(.98*4.39)/2=1.63+2.15=3.78 and (2*1.29)/2+(2*2.55)/2=1.29+2.55=3.84. Check 45 degrees (.98*3.26)/2+(.98*4.37)/2=1.6+2.14=3.74 and (2*1.22)/2+(2*2.51)/2=1.22+2.51=3.73. 45 degrees is where the second long lever comes into balance and the first long lever comes into balance at 71 degrees.

Check 38 degrees so (2.84*.98)/3+(.98*3.68)/3+(.98*4.53)/3=.93+1.2+1.48=3.61 and (2*.76)/3+(2*1.7)/3+(2*2.23)/3=.51+1.13+1.49=3.13. The third long lever has not gone far enough. Check 40 so (2.96*.98)/3+(.98*3.78)/3+(.98*4.56)/3=.96+1.23+1.49=3.68 and (2*.9)/3+(2*1.81)/3+(2*2.75)/3=.6+1.21+1.83=3.64. 40 degrees is close enough. The third long lever comes into balance at 40 degrees the second long lever comes into balance at 55 degrees and the first long lever comes into balance at 81 degrees. Only 9 more degrees for the first long lever and we can complete the task.

Check 35 degrees so (.98*2.64)/4+(.98*3.27)/4+(.98*4)/4+(.98*4.6)/4=.65+.8+.98+1.13=3.56 and (2*.56)/4+(2*1.22)/4+(2*2.07)/4+(2*2.81)/4=.28+.61+1.03+1.4=3.32. Fourth long lever has not gone far enough. Check 36 degrees so (2.71*.98)/4+(.98*3.32)/4+(.98*4.04)/4+(.98*4.61)/4=.66+.81+.99+1.13=3.59 and (2*.63)/4+(2*1.29)/4+(2*2.12)/4+(2*2.82)/4=.32+.65+1.06+1.41=3.44. The fourth long lever still has not gone far enough. Check 37 degrees so (.98*2.78)/4+(.98*3.37)/4+(.98*4.07)/4+(.98*4.62)/4=.68+.83+1+1.13=3.64 and (2*.7)/4+(2*1.35)/4+(2*2.17)/4+(2*2.83)/4=.35+.68+1.09+1.42=3.54. That's close enough as long as the long lever is slightly more torque than the L shaped lever. The fourth long lever comes into balance at 37 degrees and the third long lever comes into balance at 47 degrees and the second long lever comes into balance at 62 degrees and the first long lever comes into balance at 88 degrees. Only two more degrees to complete the task. I hypothesize that a fifth machine will reload the one weight and a sixth machine will be required to continue keep reloading one weight as a perpetual motion machine.

More to come later! If someone else wants to finish the math for me that would be great but I must do some chores and won't be on for the rest of the day probably.

[preoccupied]

Fifth long lever is 32 degrees to complete the task. (.98*2.44)/5+(.98*2.9)/5+(.98*3.49)/5+(.98*4.15)/5+(.98*4.62)/5=.48+.57+.68+.81+.9=3.44 and (2*.35)/5+(2*.83)/5+(2*1.48)/5+(2*2.25)/5+(2*2.83)/5=.14+.33+.59+.9+1.13=3.09. BINGO we have a winner! Torque on the Long lever's side has reached the reload point with about .35 torque to spare with only 5 machines! The fifth long lever is at 32 degrees and the fourth long lever is at 39 degrees and the third long lever is at 49 and the second long lever is at 64 and the first long lever is at 90 degrees.

But when the weight gets back up and rolls down the ramp it is not going to an identical weight set. Where do the weights come into balance? Do we need a sixth machine?

The first long lever is no longer in the equation. Lets try the now fourth long lever (was fifth) at 40 degrees so (.98*2.96)/4+(.98*3.37)/4+(.98*3.83)/4+(.98*4.39)/4=.73+.83+.94+1.05=3.55 and (2*.9)/4+(2*1.35)/4+(2*1.92)/4+(2*2.55)/4=.45+.68+.96+1.28=3.37. The fourth long lever has not gone far enough. Check 41 degrees so (.98*3.03)/4+(.98*3.43)/4+(.98*3.92)/4+(.98*4.42)/4=.74+.84+.96+1.08=3.62 and (2*.97)/4+(2*1.41)/4+(2*1.98)/4+(2*2.58)/4=.49+.7+.99+1.29=3.48. Fourth long lever still hasn't gone far enough. Check 42 degrees so (.98*3.09)/4+(.98*3.49)/4+(.98*3.96)/4+(.98*4.44)/4=.76+.86+.97+1.09=3.68 and (2*1.03)/4+(2*1.48)/4+(2*2.03)/4+(2*2.61)/4=.52+.74+1.02+1.31=3.59. That's close enough as long as the long lever has slightly more torque than the L shaped lever. The fourth long lever comes into balance at 42 degrees the third long lever at 49 degrees the second long lever at 59 degrees and the first long lever at 74 degrees. A total of 10 degree change has taken place after the weight was reloaded.

The weight is coming back and will need to go 16 extra degrees to complete the task, that is 46 degrees. Check 46 degrees so (.98*3.32)/5+(.98*3.91)/5+(.98*4.18)/5+(.98*4.46)/5+(.98*4.62)/5=.65+.77+.82+.87+.91=4.02 and (2*1.29)/5+(2*1.46)/5+(2*1.74)/5+(2*2.63)/5+(2*2.83)/5=.52+.58+.7+1.05+1.13=3.98. Oh wow just barely just barely able to reload the weight on the second time through. Does this mean it will reload the weight every time? More calculations to come! Any calculations from other people would be appreciated!!!

[Tarsier79]

Can you post a diagram of how you intend to connect the mechanisms, and how you phase them, and think about how this effects lifting the large weight back up?

[rlortie]

Thanks!

I have been waiting for someone to pop this question. I had a few ideas on how to accomplish this feat on his original but now I am at a loss.

Ralph

[preoccupied]

I can't visualize myself actually building anything like rlortie does because I have no tools or craftsman skills. But I can try to point out the obvious. If the axles were connected to each other they would contradict direction. A gear going one direction would cause another gear to go the opposite direction so a 1/1 gear ratio would be needed to change direction of the connection to the axles. The axles must connect to each other.

I don't know how to exchange the ball or weight. It's easy to visualize rolling the ball down ramp and into the hand of a lever but it's not easy to visualize exchanging the ball or weight from the hand of a lever to another hand of a different lever. I'm sure it would be easy for someone to visualize if they have craftsman skills. I guess it could be open facing inward on the side of the ramp and open facing outward on the side facing the L shaped lever so that when it is at the appropriate angle it will roll into the L shaped levers hand. I think the L shaped hand would have to have some kind of door on it to open a path to the ramp because otherwise the weight might roll off the hand when it is rising.

[rlortie]

'If man can write about it, man will eventually build it!' a quote by Jules Verne.

Preoc, I warned you about getting involved with the math first. You have redesigned your concept to match your equations, and although you may be happy with your summations that it will run, you are now stuck with how to make it operational.

As I see it, you only need one through axle driving as many mechanisms you feel it needs, each operating via a free-wheel clutch bearing.

As for the weight swap you can get ideas from age old marble machines, You-Tube has a good collection.

What I do not understand is how does the reload lever pull down while lifting the load weight. Correct me if I am wrong but you do not have your math equations working for you until it reaches the bottom. This means that the other mechanisms have to lift and reload while also lifting drive weights.

Gearing will not help you as no matter how many gears or ratios you use, what ever goes in friction says less will come out. All gears do is change time for movement or speed verses mass.

Building a machine that can relay the weight back and forth can be overcome, but getting the levers to meet at the bottom for roll over is another story, one that has not been written yet.

One vision I have is possibly a counter shaft such as in a standard auto transmission for low and reverse gears.

Ralph