Drawings by preoccupied
Moderator: scott
- preoccupied
- Addict
- Posts: 2026
- Joined: Fri Apr 29, 2011 3:28 am
- Location: Michigan
- Contact:
re: Drawings by preoccupied
There are two parts, the ball system on the rim and the box system in the center. The purpose of the outer ball system is to pull the lever on the bottom left by pushing on the yellow line. The lever there pushes the weight to the center in the box on the bottom left. When the box is unbalanced the ball system should be mostly balanced and it will turn 45 degrees until it is again balanced. Notice that the lever with the yellow handle is pushing the same direction as the lifted object because of gears. This idea is fresh. Is it good?
- preoccupied
- Addict
- Posts: 2026
- Joined: Fri Apr 29, 2011 3:28 am
- Location: Michigan
- Contact:
re: Drawings by preoccupied
I think this drawing looks less complicated because the colors show where the different parts are. On the bottom left box the colored arms show the different positions it will swing. The pink ball system is meant to only push on the lever on the bottom left box. In this drawing the lever is attached higher but it is still connected to the weighted arm and turns the same direction as the weighted arm because of gears. The lever on the bottom left box is colored blue where it would start orange while it's being pushed and green where it will end.
- preoccupied
- Addict
- Posts: 2026
- Joined: Fri Apr 29, 2011 3:28 am
- Location: Michigan
- Contact:
re: Drawings by preoccupied
I have two questions related to a drawing for a swastika design. The force to move a weight held by a string horizontally and a lever near vertically is less than distance positions on a lever because of the angles. How much weight is necessary to balance picture questions #1 and #2. #1 a weight is hanging on a pulley. #2 the weight is hanging on a lever connected to a pulley at its pivot. I really think it's lighter to pull on the weight horizontally by a string if it is sitting near horizontal on a lever than to try to lift it with levers but I could be wrong.
re: Drawings by preoccupied
Gday
The answer is the same as usual. There are a number of ways you can calculate this, but the universal way is probably the best.
PE = mass x height x gravity. PE will always lessen in a gravity/leverage configuration.
If you want to lift the mass on the left: All you have to do is increase the mass on the right until you are losing more PE than you are lifting on the left.
Hint: The answer will take the application of geometry to solve.
Cheers
Kaine
The answer is the same as usual. There are a number of ways you can calculate this, but the universal way is probably the best.
PE = mass x height x gravity. PE will always lessen in a gravity/leverage configuration.
If you want to lift the mass on the left: All you have to do is increase the mass on the right until you are losing more PE than you are lifting on the left.
Hint: The answer will take the application of geometry to solve.
Cheers
Kaine
re: Drawings by preoccupied
Draw an exact plan with all measurements, and I will try to hlp. You have to understand the underlying principle for yourself though.
- preoccupied
- Addict
- Posts: 2026
- Joined: Fri Apr 29, 2011 3:28 am
- Location: Michigan
- Contact:
re: Drawings by preoccupied
I want to know how much weight x can be applied until the two weights are balanced. I think the angle of the lever and the horizontal string is an important part of the calculation.
re: Drawings by preoccupied
preoccupied
A falling weight always have more leverage compared to the weight on the pole which distributes it's weight to its ground mount attached by a cable. For instant a falling weight off the edge of a table over a pulley and move a lot more weight across a level surface. I did one test with weight on wheels and I could not get enough weight on the wheels to stop the 1lb weight from pulling it to the edge stopped only by the pulley and it was close to 40lbs for the test.
At least I hope this helps
A falling weight always have more leverage compared to the weight on the pole which distributes it's weight to its ground mount attached by a cable. For instant a falling weight off the edge of a table over a pulley and move a lot more weight across a level surface. I did one test with weight on wheels and I could not get enough weight on the wheels to stop the 1lb weight from pulling it to the edge stopped only by the pulley and it was close to 40lbs for the test.
At least I hope this helps
"Our education can be the limitation to our imagination, and our dreams"
So With out a dream, there is no vision.
Old and future wheel videos
https://www.youtube.com/user/ABthehammer/videos
Alan
So With out a dream, there is no vision.
Old and future wheel videos
https://www.youtube.com/user/ABthehammer/videos
Alan
-
- Addict
- Posts: 2568
- Joined: Thu Apr 18, 2013 3:29 pm
- Location: not important
- preoccupied
- Addict
- Posts: 2026
- Joined: Fri Apr 29, 2011 3:28 am
- Location: Michigan
- Contact:
re: Drawings by preoccupied
I think I am going to make a fool of myself too
I think the horizontal string equals 1 if it really is horizontal and I think the lever gets a bonus so if it is 45 degrees it would be this bonus I think 1/(sin45)= about 1.4142
For my first question in this thread about pulleys lifting a weight at two angles the answer was 1/(sin) described by Jim_Mich to me. I really don't know if the lever is that same as that though. But it's my best guess.
If my guess is right then the 45 degree lever would make a 1.41 weight on the lever be balanced by a 1 pound weight on the pulley. And if I go towards 0 degrees instead of 90 degrees a 1/(sin22.5)=2.61 and a 2.61 weight on a 22.5 degree lever facing up would be balanced by a one pound weight. I don't think this is the solution though! I think there is much more leverage. Tarsier79 what formula do you have?
I think the horizontal string equals 1 if it really is horizontal and I think the lever gets a bonus so if it is 45 degrees it would be this bonus I think 1/(sin45)= about 1.4142
For my first question in this thread about pulleys lifting a weight at two angles the answer was 1/(sin) described by Jim_Mich to me. I really don't know if the lever is that same as that though. But it's my best guess.
If my guess is right then the 45 degree lever would make a 1.41 weight on the lever be balanced by a 1 pound weight on the pulley. And if I go towards 0 degrees instead of 90 degrees a 1/(sin22.5)=2.61 and a 2.61 weight on a 22.5 degree lever facing up would be balanced by a one pound weight. I don't think this is the solution though! I think there is much more leverage. Tarsier79 what formula do you have?
- eccentrically1
- Addict
- Posts: 3166
- Joined: Sat Jun 11, 2011 10:25 pm
- preoccupied
- Addict
- Posts: 2026
- Joined: Fri Apr 29, 2011 3:28 am
- Location: Michigan
- Contact:
re: Drawings by preoccupied
eccentrically1, I think you are saying the opposite of what AB_Hammer said.
The question is how much weight is necessary for the two weights to balance. The scenario is there is a weight hanging on the pulley - and the string is horizontally attached to a weight sitting on a lever that is facing up. When the lever is vertical at 90 degrees (Or 0 degrees at a different perspective) any amount of weight will initially pull the weight because it would be 0 space from the center of the axle. Only friction holds things at zero space from the center of an axle. I don't know eccentrically, why you would say that and maybe we are not on the same question - because I think force is distributed to the lever, just as if something were sitting on a table like in AB hammers example.
The question is how much weight is necessary for the two weights to balance. The scenario is there is a weight hanging on the pulley - and the string is horizontally attached to a weight sitting on a lever that is facing up. When the lever is vertical at 90 degrees (Or 0 degrees at a different perspective) any amount of weight will initially pull the weight because it would be 0 space from the center of the axle. Only friction holds things at zero space from the center of an axle. I don't know eccentrically, why you would say that and maybe we are not on the same question - because I think force is distributed to the lever, just as if something were sitting on a table like in AB hammers example.
- preoccupied
- Addict
- Posts: 2026
- Joined: Fri Apr 29, 2011 3:28 am
- Location: Michigan
- Contact:
re: Drawings by preoccupied
If I can answer these leverage questions 1 and 2 about my swastika drawing I can then work on my more complicated leverage questions like here... This A shaped thing pulls a rod through it and pushes out its wings by two inner connections until the inner connections are aligned. I think there is leverage advantage in pulling the wings apart by pulling on the rod when the inner connections are nearing becoming flat or aligned. This idea is for the wheel design in the picture3 also, where the box and lever with weight at the end of it in the center of the wheel is always mostly balanced but the two weights at the rim are pushed apart at a mechanical advantage. That mechanical advantage has to go somewhere, so I assume it would go to centrifugal force and turn the wheel. I probably wouldn't be thinking about this idea if not for some of the writing by Jim_Mich that I've read. I haven't read very much but what I have read is interesting.
There is no way for me to know if my motion wheel in picture3 is any good unless I can calculate the leverage questions. I need a mechanical advantage when I push the weights apart to speculate if I potentially put that mechanical advantage into centrifugal force. I just am really an amateur and don't know a lot of math.
There is no way for me to know if my motion wheel in picture3 is any good unless I can calculate the leverage questions. I need a mechanical advantage when I push the weights apart to speculate if I potentially put that mechanical advantage into centrifugal force. I just am really an amateur and don't know a lot of math.
- preoccupied
- Addict
- Posts: 2026
- Joined: Fri Apr 29, 2011 3:28 am
- Location: Michigan
- Contact:
re: Drawings by preoccupied
I propose and hypothesize that oscillating two arms from the center to a right angle produces maximum impulse. Milkovic pendulum showed that impulse can move something on wheels but I propose that his oscillations could be inefficient and if they started in the center and moved out to a right angle that it might have maximum impulse. I was swinging around my arms and that seemed to feel like the area that had the most impulse, so that's why I have my hypothesis.
In my ms painting motion at right angle, I show with arrows where the centrifugal force is going and then the blue line is where the force isn't and that is where I think the impulse is. Or at least that is where I think I felt the impulse when I was moving my arms around. If I'm right this oscillation could propel ships and possibly perpetual motion machines. I won't hold my breath.
In my ms painting motion at right angle, I show with arrows where the centrifugal force is going and then the blue line is where the force isn't and that is where I think the impulse is. Or at least that is where I think I felt the impulse when I was moving my arms around. If I'm right this oscillation could propel ships and possibly perpetual motion machines. I won't hold my breath.
- preoccupied
- Addict
- Posts: 2026
- Joined: Fri Apr 29, 2011 3:28 am
- Location: Michigan
- Contact:
re: Drawings by preoccupied
To support my hypothesis I have the idea that because I stopped at a right angle the centrifugal force has to distribute evenly below and above this pink line. That would then mean that the force would not be able to distribute fast enough because there wouldn't be enough space and the extra pink centrifugal force would go towards the blue line. I have no real reason to have this idea, I just do because it looks to me like there is less space for some reason. I'm literally imagining this whole idea. Only experimentation would show reason in all of this.
re: Drawings by preoccupied
Referring to your top figure, which I have duplicated below, the angle of the string is important. So on 'both figures, you need to provide the height of the
pulley above the base (T and U). Also on the right, you need to provide the length V, as you cannot solve an equation with 2 unknowns, unless they are related in some way. Also can you specify what 45 to 90 degrees is? will the lever be horizontal, or vertical, and are you lifting the "Y" lever from hanging straight down to horizontal, or from horizontal to vertically up?
In answer to your basic question,in lifting the weight "x", #1 will be easier to lift than #2. This is due to the distance X will travel vertically in comparison.
ADD: also, the distance between the two vertical levers will also be important to determine the angle of the string. The angle of the string determines how quickly the weight on the left is lifted compared to the weight dropping on the right in both #1 and #2.
pulley above the base (T and U). Also on the right, you need to provide the length V, as you cannot solve an equation with 2 unknowns, unless they are related in some way. Also can you specify what 45 to 90 degrees is? will the lever be horizontal, or vertical, and are you lifting the "Y" lever from hanging straight down to horizontal, or from horizontal to vertically up?
In answer to your basic question,in lifting the weight "x", #1 will be easier to lift than #2. This is due to the distance X will travel vertically in comparison.
ADD: also, the distance between the two vertical levers will also be important to determine the angle of the string. The angle of the string determines how quickly the weight on the left is lifted compared to the weight dropping on the right in both #1 and #2.