Posted on

# Wiseman’s Wheels eBook Corrections

This page is intended for those people who have downloaded the ‘Wiseman’s Wheel’ pre-view eBook version 150119.

As I’ve been answering questions I’ve become aware of several issues.

1. Some ‘lables’ that need to be put on the Stage 6 drawing

S = small wheel
Sg = small wheel gear (0.2222’ diameter), securely fastened to S, sharing same axle.
B = big wheel
Bg = big wheel gear (2’ diameter), securely fastened to B, sharing same axle.

B and Bg are fixed together, S and Sg are fixed together, Bg and Sg are interlocked.
So full negative torque of B (radius of 1′) is applied to Sg (radius of 0.111′)

As far as Sg is concerned, the full weight of B weights are applied to the Sg.  In this calculation, the ratio of torque of B weights to B axle is meaningless.  This is what I mean by B ‘freewheeling’.  The B wheel (in this case) only exists to hold the weights aligned, to register back with the S wheel.

2. Number of balls on B
To clarify, there are only 4 balls on B at any given time.  As a ball is added from S12 (corrected) to B1, a ball is subtracted from B5 to S6.

3. S wheel miss-numbers
I see I mis-numbered 11,12 and 13 on Sketch 6.  (Corrected already).

4. The Stage 6 drawing needs the gears drawn in. (I’ll do that)

5. The Stage 6 concept needs clearer explaination. (I’m working on that)

6. Long math for calculating Sg radius:

D = diameter, C = circumference, R = radius, π = 3.14159
Calculating the circumference of a circle with a diameter (D x π = C).
Calculating the diameter of a wheel from the circumference (C / π = D)

Bg = 2’ diameter = 6.28318 C
To find diameter of Sg that will rotate 45° every time Primary Wheel rotates 5°
# of 5° in 360° = 360/5 = 72   6.28318 / 72 = 0.08726639’ per 5°
0.08726639’ x 8 (8 of 45° to get 360°) = 0.69813112’ C for Sg.
Sg D = 0.69813112 / 3.14159 = 0.2222’
Sg radius = D / 2 = 0.1111’

7. Stage 6 math contains a fundemental conceptual mistake
Jan. 21/15 … I made a mistake.  Not the first time and won’t be the last.

Re-Calculating S wheel torque:  http://www.engineersedge.com/calculators/levers/page_levers_1.htm
F = (W x X) / L  or  F x L = W x X
W = 3.977 lbs.  X = Sg is 0.111’ (1.332”) radius.  L = S is 0.25’ (3”) radius.
3.977 lbs x 0.111’ = 0.442 ft/lb   0.442 ft/lb / 0.25’ = 1.766 ft/lbs counter-clockwise.

Turns out that the S works like a lever balance and the ‘balance’ force the S wheel needs to overcome at it’s rim (from 3.977 lbs at Sg rim at 180°), is 1.766 lbs clockwise at S rim 0°…
NOT 0.442 ft/lbs counter-clockwise as I’d assumed in Discovery Stage 6.

So… Assuming counter-clockwise rotation, the S wheel has;
a ‘positive’ (weight lifting) torque of 1.766 ft/lbs (from Bg leverage through Sg) + 0.177 (Weight 6) = 1.943 ft/lbs and
a ‘negative’ torque of 0.177 (Weight 8) + 0.250 (Weight 9) + 0.177 (Weight 10) = 0.604 ft/lbs…
Weights S7 and S11 are neutral.  There are no weights in positions S12 and S13.
The NET positive counter-clockwise torque on S would be 1.339 ft/lbs.

So S on Stage 6 would turn counter-clockwise as the B wheel turns clockwise, just the opposite as I had ‘assumed’ with my initial Discovery Stage 6 concept/math.

I’ll get a leverage drawing made, to make this concept clearly understandable.

Balls would be fed from S12 (corrected) into B1 and from B5 into S6

I made a mistake and it SEEMS to have worked out better than I could have imagined…

8. The eBook download process is cumbersome.

I’d like to make it smoother, but I need feedback as to where people are having trouble.  Please report to me where you had issues and, if possible, make suggestions as to how I could help people do it smoother.

I can’t just send a PDF because my website is fully integrated to support my eBooks, with Resources, Forums, Affiliate stuff, etc.  The eBooks (real version, not what you have) are intimately linked to my website.

And the eBooks are each custom generated through this process, ‘stamping’ personal information on most of the pages, so people are less likely to spread the information freely (because they’d be giving out their personal information).  People who want to share are then more likely to share the download link instead, which helps my affiliate program.

Another advantage of the ‘custom stamping’ my website does is that I can easily track down people who are sharing inappropriately (by looking at the eBook they ‘shared’ on someplace like Scribd) and shut off their website access to Resources/updates/eNews/etc. (even prosecute if I choose).

This system also ‘protects’ me from people just spreading my literature virally without compensation to me.  Anyone who gets a version that isn’t registered to them is limited in it’s value, because they can’t access the online Resources, etc.  The ‘viral’ version then encourages them to register.  I don’t patent my innovations, so this (selling information and helping people replicate my work) is my way of making a living… I NEED that income to continue to exist.

9. Dave Brown points out that Sketch 5 has an issue
in that the inner weights don’t line up precisely with the outer weights:

The weight #67 from the inner track would leave when the inner track angle is 36.9705° to instantly (horizontally) meet weight position #1 which is at 17.5° on the outer track.

I get that.  It doesn’t change the torque because there is no longer a weight at #67

But, I see I got the angle on #8 wrong.  It should have been 342.5° (-17.5°).  It doesn’t change the math at the end but accuracy is important.

The weight #8 from the outer track at angle -17.5° to instantly (horizontally) meet weight position #9 which is at -38.03° (321.97°) on the inner track.

I get that too, thank you.

That would reduce the torque of the wheel by about 0.017 ft/lbs and make it more difficult to ‘register’ the two tracks together…

Have to think on that one.

1. Maybe have the weight on the outer track leave early. What would that angle be?) to meet #9 at 325°

2. Make the path to the inner track raised a bit? (3° would cost us a bit of power but we’d gain some torque to compensate).

3. Change the geometry, (try to find a place where the inner and outer tracks line up)?

4. Rotate the inner weights (like I did the outer) so that they more closely line up horizontally.

5. Go to a secondary wheel, like concept 6?

OK, I ‘see’ something.
We may have to rotate the inner weights a couple degrees but then we can take advantage of the horizontal  ‘miss-alignment’ geometry. I think this idea can work as long as the total exchanges take place within the designed 5°.

We can take advantage of the slight misalignment between the inner and outer tracks by NOT having instantaneous ball exchanges.

We can stagger when we apply the weight exchanges and thus we can gain more torque.

This idea presumes we have a horizontal ‘track’ (shaped like a half donut) that is ‘fixed’ (separate from the wheel) and held horizontal.  It is sized to ‘register’ with the inner and outer weight holes, as they com into alignment with it.  We have such a track on the top and bottom for ball exchanges.

So we add a weight to the inner ring #9 (from the horizontal track) slightly BEFORE we take the weight from #8 on the outer ring…
And we add a weight to #1 on the outer ring (from the horizontal track) slightly BEFORE we take the weight from inner track #66.
So for a couple of degrees we actually have an additional 1+ ft/lb of torque.