Progress towards the Screw Wheel

[beapilot]

This thread is made to make this Screw Wheel possible. It requires a lot of calculations and different settings to get different results. Since this became toxic to my time ever since it came to me in a dream, I decided to seek help to the community.

After I did hours of tinkering, I still ran into questions.

How the wheel works:
The screw will go up as the wheel turns. The rod is made so only the resisting force is the direction of where the wheel is spinning, to the right which means there will be a back force to the left.

Things to have in consideration:
Screw angle
Load
Screw diameter
Required force to overcome back force or static force

My goal right now is to develop a generator that will allow inputs to be placed in to search for outputs on an instant. Example in the attachment.

Attached is a document where I can plug in the screw diameter, screw angle, and the load. Then I will simply look at the yellow column under the first chart. If it is a negative, it means that the screw will stall or or go to the left. If positive, then it will not move to the left or stall.

According to the data, I concluded new questions:
When the screw angle is 89 degrees, then when it is at zero degrees with the wheel, then the force is not zero. So, I needed to figure out what is the force required to push the female screw to the right.

Then I wanted to check and see if I can move the stationary rack higher. If I do this, I can add a larger circumference gear but will not do much difference.

I know that I can max have three rods touching 1 quadrant.

Now, it is not hard to reverse the screw. Just have the rack on the other side of the wheel which will cause a reverse back spin. It is important to fix the problem with 1 quadrant first.

So, just look into the document and check out how I am formulating all of this.

Then I thought have the screw angle change as the female screw gets higher.

I really did not want to type all of this wasting so much time getting all of this neat and all. Then just to find out the wheel will not work after all.

Just because I know Trigonometry does not make me a master at it. If I have a mistake, please help me out.

Lets make this possible!

In the document, play with the purple fields. Type in 89 degrees for the screw. Check out the difference. Try 85.

Have fun!

[beapilot]

Instead of using the female screw, a block screw can work. Basically same process but the male screw threads will interlock with the block threads. The block holds the green dot (shown at top but should be centered) that goes to the other side of the wheel. From there the load will hang.

Bearings such as wheels and other things to support the slide and screw thread surface has been tinkered of.

[path_finder]

Dear beapilot,
Your design concept is interesting. Following your demand here are my first comments:

Basically you want to obtain a translation along the radius.
A screw assembly (male + female) is obviously the way to get the most important friction losses.
IMHO a closed Archimedes screw filled with mercury shall be much more efficient for the same purpose.
The conical gear at the center will give you the desired ratio for the twist of the rod, but will increase again the friction.
There are some other mechanisms available for this kind of radial translation, an half sized circle acting as an hypocycloid being one of the most known.
I'm not sure the calculus could be of any particular use in the specification of this kind of design, the most important being mechanical.
Many thanks for sharing.

[Tarsier79]

So why is your design different to the diagram below.

Pathfinder makes a good point... But this has obviously been tried before.

Why would a screw driven lift take less energy to lift a mass than with a lever. EG. If the screw lifts at a20% gradient, it has to travel 2 times further than a screw lifting at 40%, but at half the effort, just like a standard lever, correct?

[beapilot]

Path-finder & Tarsier79,

Thank you for your feedback.

I can only say that the Screw Wheel design is different. I can not say it works.

Why the Screw Wheel design is different:
It is a complicated form starting at 90 and ending at 0 degrees.
As the screw lifts the load at 90, the load becomes lighter as the wheel turns the screw. As the wheel turns, the height increases making the torque stronger while the weight of the load increases on the wheel.

-------

Progress report:
This morning as my mind came to ease with slight pulsing on my head, I was able to think clear again.
When the 89 degrees screw is resting horizontally at 0 degrees on the wheel, it requires a lift of (Assuming the load is 5)

5*SIN(89)= 4.999

When that happens, it will attempt to slide along the 89 degree incline. The amount of force moving down it is:

5*COS(89) = .0872.

I will place this in the generator while attempting to figure out how to make it calculate with the other data smoothly.


Joshua


Edit:
When using the generator, the brown box below the brown box above it shows the final diameter of the wheel in feet. Lets think realistic now.

[beapilot]

Fixed formula:

(Load)*SIN(((360*PI/180)-((90*PI/180)-(Screw angle))+(Angle on Wheel)))

This shows the amount of force on an incline with the input of a load, screw angle, and where the screw is on the wheel.

[Tarsier79]

It seems to me you need 2 formulas

1. the amount of force your mechanism is producing.

2. the amount of energy it takes to push the weight into its position.

Then put them into a spreadsheet to solve for 360 degrees....

[beapilot]

Tarsier79,

Thank you for your opinion. Number two comes before number one.

This wheel requires a lot of different formulas to find out the total force the wheel can produce on its own.

Currently I am in progress coming up with better formulas and fixing existing ones in the generator spreadsheet.

Joshua

[beapilot]

Hello,

Attached is a newer version of the Generator for the Screw Wheel.

Feel free to play with it.

I would appreciate it if someone looks this over to help me come up with flaws.

Many Thanks,

Joshua

[beapilot]

Now I can come to a rest. YES!

Below are two attachments. SW Doc is a document that explains some steps to where I caught a pretty ugly error.

The other attachment is the latest progress of the generator up to date including fixing the ugly error.

Sadly, if the fix is right, you will see in the result yellow column.

I saw 11 downloads of the previous version of the generator and no feedback.

Thanks.

EDIT:
I am finding anything useful this wheel can have. I currently found height ratio. If you find anything that could be an alternative, post away.

[ruggerodk]

Dear beapilot,
I can't figure out your equations...why, I'm not so much into math physics.

What I see from your drawing, is a way of weight displacement into a larger radius; from zero displacement at 12-o'clock (90°) through an accumulated larger radius during the whole structure's 'fall' to 3-o'clock where the weight will reach the optimal largest radius.
I also see, that the 'inner gearing' - that turns the threaded rod - has some kind of tension/friction....and too between the threaded rod and the cart.

My own experiments of similar displacement but by different application, show that it is indeed possible to have this kind of displacement as soon as the weight passes 12-o'clock.

What I do not understand is why you have to include the balanced (opposite) weight into the equation..?

To me it is straight forward: At any point the weight's tendency to fall is greater than the resistance to displace itself to a larger radius...and this 'preponderance' accumulates more and more during the 'fall' from 12 to 3-o'clock.

regards
ruggero ;-)

[beapilot]

ruggerodk,

Thank you for your reply.

I want to make sure I am understanding what you are trying to say.

Here is your question:

What I do not understand is why you have to include the balanced (opposite) weight into the equation..?

To me it is straight forward: At any point the weight's tendency to fall is greater than the resistance to displace itself to a larger radius...and this 'preponderance' accumulates more and more during the 'fall' from 12 to 3-o'clock.

From the information provided, I think you are trying to say is:

If a distance of a lever is longer than the opposing distance, then a greater torque should occur.

That is right.

On a wheel, there is usually an opposite weight. If 8 weights, that makes 4 lines or rods. If all weights are equal distance, everything is balanced and the wheel does not turn. Let us take one rod and call it a lever.

Let us assume there is only one weight at one end of the rod and it rest in the first quadrant which is in your terms from 12 to 3o clock as in 90-0 degrees. If I have a 5 mg, then a portion will be absorbed on the stationary axis and become useless to the entire operation of the wheel and adds friction. The document should help clarify that.

Now, why have a weight on both ends? I need a way to lift the weight back up some how and keep the wheel in neutral position. So, as the height of one weight increases, a greater torque appears while the other weight do not increase in distance (To make things simpler to understand but all the weights will shift iff quadrant 1 works). So I had to take care of that. Now, you might be thinking, oh, so we have a larger distance etc but the total torque does not change. That is because I am focused at one point of the rod which is the distance from the axis to the rack which I called the Small Wheel.

So, the equation of a lever system is F1D1=f2d2. Lets say I want to find the amount of available lift force I can do by the given current distance and weight of the rod. So, assuming you read the document, you see that you can lift MORE mg but at a smaller distance. Now, because we knew all along that that lift MORE includes the balance weight. So I have to subtract it to get the REAL USEFUL lift weight.

If I did not clarify your question, please rephrase it.

EDIT:

Check out the image to help clarify what I am trying to say.

[ruggerodk]

Dear beapilot,
I'm not sure I fully understand your answer - but I really do appreciate your effort in trying to explain it to me. Sorry ;-)

Now, why have a weight on both ends? I need a way to lift the weight back up some how and keep the wheel in neutral position. So, as the height of one weight increases, a greater torque appears while the other weight do not increase in distance (To make things simpler to understand but all the weights will shift iff quadrant 1 works). So I had to take care of that. Now, you might be thinking, oh, so we have a larger distance etc but the total torque does not change. That is because I am focused at one point of the rod which is the distance from the axis to the rack which I called the Small Wheel.

It still confuse me when incorporating any other opposing weight, torque or whatever into the most simple (imho) lever application: a falling weight-rod/stick.

In my scenario I would prefer setting the F1 and D1 to zero - the reason being not to complicate things more than necessary. My first and only concern is the weight's displacement from the (zero) fulcrum (or pivot point). That way I can concentrate on two major 'events': The weight's tendency to fall and its resistance to 'crawl up' (and out) the rod to a larger radius.
The ratio between those two, to be precise.

How to - as you say - "lift the weight back up" is not an issue at this point...at least not to me.

I hope you understand where I'm heading - otherwise I am a graphic designer and it will not harm me to make a drawing to support the wording.;-)

regards
ruggero ;-)

[beapilot]

ruggerodk,

I am pleased to hear you are a graphic designer. That means you have the proper tools to draw on the computer. When I drew that lever image, it took me 45 minutes with only a mouse and MS Paint, yuck.

It seems to me that you are building a design based off this concept because there are things you are not concerned about.

If I did not need to worry about the weight to go back up the wheel, then I would have a 100% PMM.

I NOW understand where you are coming from. You only want to work with ONE (1) weight in the first quadrant assuming there is no other force besides friction.

If you would like that to happen, play with the generator below and check out some awesome results in the yellow column. All I did was took out the opposite weight or balance weight. Do not take this current generator attached as the real results.

If you please, do help me understand your idea on an image. I am a great visual learner.

Joshua

[ruggerodk]

Dear Joshua,
I have just read you post - and looked at the yellow column in tyour spreadsheet....Im not sure what the figures tells me. Though it looks like there are a gain in force from 70 down to zero degrees.
I can't see any figures of the displacement...am I misreading it?
Perhaps you could explain what I'm supposed to see, please.
In the meantime I will make a drawing to show you what I mean.

regards
ruggero ;-)