Idea I had since elementary school

[preoccupied]

The L shaped lever when the L shaped part is at its longest is 2.83 from the fulcrum. The 2.83 lever is then 45 degrees and is 2 from the fulcrum. 1.4*2.83=3.962 and 2*2=4 so at the longest point on the L shaped lever 1.4 weight will make it rise. If the 2.83 lever is lowered 3 degrees then 2*1.89=3.78 and 1.4*2.826=3.956. So I made another mistake in my calculations! I thought that because the longest point on the L shaped lever was the strongest that that determined if the L shaped lever would lift to the ramp. I was wrong! Lets start with 1.3 now 2*1.89=3.89=3.78 and 1.3*2.826=3.67. Lets go 10 more degrees so 2*1.5=3 and 1.3*2.75=3.57. 1.3 won't do it! Check 2*1.5=3 and 1.2*2.75=3.3. Check 1.15*2.75=3.16. Check 1.1*2.75=3.03. Check 1.05*2.75=2.88. Lets go 10 more degrees so .92*2=1.84 and 1*2.6=2.6. Lets go back 5 degrees so 1.28*2=2.56 and 1*2.69=3.69. Lets go back 1 degree to 17 degrees so 2.7*1.1=2.7 and 1.33*2=2.66. Lets go one degree back to 16 degrees so 2.72*1=2.72 and 1.37*2=2.74. That's about right. 45 degrees plus 16 degrees = 61 degrees. The height the L shaped lever can reach if the weight is 1 pound to 2 pounds is .78 inches. So for 2 inches + .78 inches 2.78 inches is the length the Long lever has to fall. So the new length of the long lever is 5.56 because the degree is 30 because I am making a 7/10 gear ratio. The long lever is longer but the distance it travels is shorter however the gear ratio is the same. I don't know how this will effect the calculations but now I have to do the calculations all over again. I wish I didn't have to do guess and check. Would someone with math education do this without guess and check for me? With the right math equation the calculations would take less than 5 minutes but I have to do guess and check over and over again because I don't remember my math education from High School and it takes forever.

[preoccupied]

My math was wrong again. The degree is from 90 degrees. 90/30=3 so for every one turn the other gear needs to turn 3 times 1/3 gear ratio. And if I do 90/60=1.5 so for every one turn the other gear needs to turn 1.5 times 2/3 gear ratio. I don't know where I got .7 gear ratio from that was silly of me. 4.81/3=1.6 and 5.56/3=1.85 and 1.85+1.6=2.45<5.66 so a weight should not be able to be reloaded. Check 1.6/1.5=1.06 and 3.21/1.5=2.14 2.14+1.06=3.20<5.66 so there is no chance in hell that will reload a weight. 90/80=1.125 so .49/1.125=.43 and 2.83/1.125=2.51 2.51+.43=2.91 so there is no chance to reload a weight. 90/10=9 so 15.76/9=1.75 and 16.009/9=1.78 1.78+1.75=3.53 so there is no chance to reload a weight. I think as the long lever gets longer and the gear ratio gets higher it gets closer and closer to breaking even but won't reach the break even point. I created a HOAX! I was wrong because my math was wrong. Oops. No weights can be reloaded and that's a darn shame.

[Tarsier79]

See the below diagram. This math is perfect world, and assumes no lever weight.

I have used a 4:1 weight ratio, and lever lengths of 1,1,1

Using Trigonometry only to calculate the leverage force each weight applies...

The COM, will be 1/5 lenth of "D" from the lower weight, (1/(1+4)), or a distance of .447.

When this hangs straight down, and both weights are balanced, the angle will be 18.434 degrees CCW from the position shown. So to swing it up to this position using inertia, its starting angle will be 36.868 CCW.

The heavy weight will travel a height of .2, the light weight will travel .8 (funny that...)

The small weight magically rolls horizontally 100, but can only fall .8 to be reset, so has a fall angle of .458 degrees, giving us a gear ratio of 1 : 80.498(close enough to 80.5) In this position the small weight applies a leverage force of 1.25 (1x100/80.5 ), the large weight hanging at 90 degrees at that exact moment will be applying 0 rotational force.

At the point the small weight needs to be to reset, it will be a distance of 99.997 from the axis /80.5= lifting force of 1.242. At this point the large weight will be horizontal distance of .8 therefore 4x.8 giving us a leverage force of 3.2

So at the bottom of the movement, the large weight will be driving the small one back upwards...But thats not the entire story

At the start of the long lever movement, the small weight is applying more torque than than the large one, and due to its lever length/ gear ratio, also has more inertia than the heavier weight. Although we cold find the balance point of these two mechanisms, finding where they will swing to will be much more difficult, and is beyond my capabilities. Logically, what you will find, is that in this perfect world scenario, the small weight will drive exactly to the reset point. Why?, because as I stated previously, discounting all trig functions above the simple formula is weight x height=weight x height.

Moving to real world.... the ball needs to roll downhill to achieve horizontal distance, friction is a .... all levers can be balanced, so their weight (apart from inertial properties) can be mostly discounted.

Mathematical models are fine, but dont always tell the entire story, and as you see, mistakes can be made. I recently contacted a person regarding their water driven gravity device, and according to his calculations, and initially also mine, it was a runner... Until you realise the mistake. How did that extra leverage get there? Sometimes you just have to build something to get the final answer.

BTW, did you end up doing your swastika vs straight lever torque test? And do you still believe a right angled lever gives an advantage?

[preoccupied]

I still pan to test the torque on a right angle bar. The guy with the torque wrench hasn't found his torque wrench and might not have the torque wrench type that measures torque but instead the kind that clicks into place when a certain torque is reached. I might have to buy a torque wrench. I think the right angle bar creates mechanical advantage because that's what it seemed to do with a car jack I was using. Pushing towards the inside of the right angle bar created more torque than pushing towards the outside of the right angle bar it seemed. It seemed like pushing towards the outside was more difficult than normal as if it were anti mechanical advantage.

[rlortie]

I have never understood this car jack scenario? pushing and or pulling towards the inside verses the outside of a right angle.

Could you be a little more explanatory about this, you have been asked before of what kind of jack you speak of. I have never seen an answer, yet you keep comparing to it.

The only thing that makes sense to me is that you cannot push down on a horizontal lever any more than you weigh, when your feet leave the ground that is the end of applied force. If your feet do not leave the ground then you are pushing with less than your weight.

If you are in average physical condition for a person of your age you can lift or pull up more than your own weight. pulling up on the jack handle is always easier than pushing down. IF you can chin yourself on a chin-up- bar then you can lift more that you can push.

For this reason I have I always felt that a Handy Man Jack was designed upside down, you have to push down to lift the load.

http://www.amazon.com/s/ref=bl_sr_autom ... in=Hi-Lift

Ralph

[Tarsier79]

Well, for a start, you can see the right angle provides no mechanical advantage in the above scenario.

An alan key provides mechanical advantage through a right angle, but this is purely leverage, as in the case of your car jack.

[rlortie]

Why do you repeat telling us that your friend has not found his torque wrench, of what use will a device set to recognize a preset value be to you! You will still have to build the machine first to apply it, only to find that unless it a torque wrench with a read out meter it will be totally useless.

If you read all ongoing threads on this forum you would have recently seen the answer to your need of torque values by clicking on;' http://my.voyager.net/~jrrandall/PronyB ... Brake.html
Located in 'Making a platform for your gravity wheel designs'

A torque wrench uses the same principle as the above mentioned alan key. Only the wrench requires a socket applied at a right angle to the ratchet handle.

Ralph

[preoccupied]

give me a link to an alan key. That sounds interesting.

[rlortie]

Must we do everything for you?

Learn to use Google; type in Alan key or Alan wrench. You will quickly have your choice of many explanations.

Edit: The proper spelling is 'Allan' wrench Patented 1943, but over the years one 'L' has been dropped. a search using either one or two 'L's will bring the same results.

[rlortie]

I have made a great discovery, it's Redneck technology! Bessler's secret revealed, a scissor jack with a right angle crank. I did not know that Bessler was a Redneck!

preoc, is this the type jack you are referring too. I and others have asked many times, you refuse to answer.

This one turns a lot easier counter clockwise than it does clockwise, I wonder why?

Picture supplied by Redneck technology 'I can fix anything"

Edit: More clues; It is easier to push down on the left than the right. It is easier to pull up on the right than it is on the left.

Ralph

[Richard]

Ralph ask

This one turns a lot easier counter clockwise than it does clockwise, I wonder why?

This clearly being a rhetorical question, as to your understanding...still needs an answer.

You seem to draw attention to the fact that torque is applied differently because of the direction it is applied.

being applied (counterclockwise) you need no extra force..you need only to control the descending (mass / weight)

being applied (clockwise) force must be added to lift the ( mass / weight).


Do we infer, from you...that we should not exclude, controlling the descending mass / weight..in the design?

[rlortie]

Richard,

You need not infer anything from my rhetoric questions. It is rather obvious that I know why more torque is needed to lift than to drop.

The inference (rhetoric) is an attempt to use language effectively and persuasively hoping to seek some direct answers from preoccupied.
His explanation of his use of the jack, makes me wonder if he is aware of this factor.

True, you should not exclude, controlling the descending mass / weight..in any design. without control and just letting it drop you gain nothing. I need not question your ability to know that. But this subject is not the intent for this specific thread and circumstance. Consider my statement as a 'pun', a play on words.

Ralph

[preoccupied]

I don't know what kind of jack it was. I was able to pull into the right angle or out of the right angle going the correct direction (pushing the car up). The right angle could flip back and forth at the axle so I could handle it at different angles. When I angled it so that I pulled or pushed into the right angle it was easy. But somehow it was difficult pushing or pulling outside of the right angle also.

[rlortie]

preoccupied,

OK! Now we are getting somewhere ( I hope)... I believe I now know what you are talking about.

Where the handle connects to the Jack screw is not unlike a simple universal joint. When in rotation at on off angle it is forced to turn faster 1/2 the revolution to compensate for the slower !/2.

This is why automobiles have two universal joints, on front and one rear. They are installed 180 degrees from each other which cancels out the pulsating speed. see Universal joint: http://en.wikipedia.org/wiki/Universal_joint

In most of the older luxury autos the universal joint was replaced with a Constant Velocity joint also known as a double Cardan;
http://en.wikipedia.org/wiki/Universal_ ... rdan_Joint

Within these to links you will find the math that explains why you found the jack handle to require more force when you did not have the crank at a right angle, or in line with the jack screw.

Ralph

[rlortie]

OOPS!

I pulled a boo-boo in the above post

This is why automobiles have two universal joints, on front and one rear. They are installed 180 degrees from each other which cancels out the pulsating.

This should read '90 degrees not 180 as 180 in a 'U' joint puts you right back where you started.

Ralph