Previous Update....... Updates Index.......My Post-Trib Book

## TRACKING ANTI-CHRISTIAN NEWS

December 26 - 31, 2023

Perpetual-Motion Wheel -- Watts

### Grappling With a Momentum Factor

Here's a drawing you'll need for this discussion: Vertical-Spoke drawing:
http://www.tribwatch.com/photos/perpetualWheelLines.png

Although I'm calling the liquid, water, it should be anti-freeze.

To find total power advantage on one side of the perpetual-motion wheel I've been laboring with in the last few updates, I think we need to combine gravity bite with momentum, but having struggled with this task for days, I don't think one can put the two into one formula. The attempt is making the wheel more familiar to me in spite of the confusion I'm in as I tread on territory not familiar to me.

Gravity bite involves inches between the water on the wheel and the vertical axle line, and momentum increases with increasing inches between the water and the central axle. However, gravity bite never changes while momentum changes with changing velocity. It's giving me a hard time trying to figure out how to make a formula, and whether to trust it, for the sub-purpose of finding the advantage of the do-work half of the wheel. The purpose of the sub-purpose is to find how much power this wheel might have, and pass it on to anyone who might want to build a wheel.

It's not fully true that gravity gets more bite on the tubes that have more distance from the axle. Tube 4, in the vertical-spoke drawing, has water far from the axle and yet has almost zero gravity bite due to its position at the bottom of the wheel. This is why we might measure gravity bite to the axle line, not the axle, for Tube 4 is very near that line. Tube waters closest to the axle line therefore get the least gravity bite.

[Insert January 1 -- However, one can make the argument that measuring to the axle likewise makes Tube 4 of little wheel-turning effect. That's because any reduction of its axle distance keeps it roughly on the same sharp angle, and angle of the tubes, as measured from the axle, has half to do with calculating force levels in creating torque. In this update, I did all the calculations with inches from the axle line, but I think I will also re-do the calculations with inches from the axle. End insert]

What advantage does Tube 2 have due to its far distance from the axle? When the wheel is moving, this tube gets more velocity than its counterpart, and by the same amount it also gets more lever power, called torque, for turning the wheel, though torque applies only on the right side of the wheel. The left side applies counter-torque as gravity pulls the water down counter-clockwise.

On the other hand, this is tricky because momentum applies torque too. Momentum makes the water "heavier" on the way down, yet momentum exists also in the upward swing. This calculator tells that, when this wheel is going one revolution per second (31.4 feet per second), each pound on the wheel grows to about 1.98 pound-force, because it carries .98 pound-force above and beyond its own weight. It almost doubles in weight.

Upward-moving tubes get gravity bite, but have no momentum added to them directly by gravity. The wheel turns because gravity pulls do-work-side tubes more strongly than go-against-work tubes, thanks to the advantage, or call it the remnant force. The momentum of upward moving tubes is sourced directly in the do-work-side tubes.

This work is not far from finishing because we need only to calculate 1/8th turn, because that bit of turn will be repeated identically and perpetually. We don't need to separately calculate every 1/16th turn around the wheel. Once we've nailed down the correct details of just 1/8th turn, we're done with the sub-task.

As per the position of the wheel in the drawing with vertical spokes, the water around the wheel is essentially in balance as far as inches to the axle line are concerned. This gravity-bite factor has two faces, however, as I explained in the square brackets of the last update: 1) distance between water and axle line; 2) angle of the water from the central axle. Gravity turns the wheel by applying itself to both of these facets.

As soon as the wheel starts to spin from its vertical-spoke position, the water slides fully down both Tube 1 and Tube 5 to begin giving the do-work side it's largest gravity-bite advantage too.

Although the gravity bite all-around is almost equal if we include only the inches from the axle line, the do-work side gets an advantage when we combine those distances with the angle factor. The latter modifies the true distance, making it as though it were a lesser distance, wherever the angle is less than 90 degrees (occurs for all tubes). The numbers below are as per the drawing above:

Do-Work Side, VERTICAL SPOKES

Spoke 1: 18.6" x (27-degree / 90-degree) = 5.6 inches of force (max 60)
Spoke 2: 48.6 x (85 / 90) = 45.9
Spoke 3: 39 x (50 / 90) = 21.6
Spoke 4: 6 x (8 / 90) = .5
Total Side inches of force = 73.6 (maximum 240)

Go-Against-Work Side

Spoke 5: 18.6" x (27 / 90) = 5.6 inches of force (max 60)
Spoke 6: 30.6 x (57 / 90) = 19.4
Spoke 7: 39 x (80 / 90) = 34.6
Spoke 8: 21 x (32 / 90) = 7.5
Total Side inches of force = 67.1 (maximum 240)

While the total, real inches to the axle line are about the same for both halves of the wheel, the angle factor alters the true numbers and gives a solid advantage, 73.6 to 67 inches, to the do-work side. The surplus or advantage in percentage is 73.6 / 67.1 = 1.1 = 10-percent larger. After cancellation of all tube inches on the anti-work side, the do-work side has this 10-percent larger-inch status by which to turn the wheel. But I feel safer using the inch surplus instead, to find how many watts this wheel can make.

The "inches of force" here refers to torque. Don't try to imagine anything such as a force pushing/lifting over 1 inch of distance. It doesn't mean that. The 8 different inch results are the new water distances to the axle line, yet while they are not the real distances, they represent the real amount of torque force. I'm using the inches as a representation of torque force. The higher the inches, the more torque, all proportional.

See what happens very shortly after the vertical-spoke position, after the full transfer of water (see drawing here):

Spoke 1: 32.4" x (42-degree / 90-degree) = 15.1 inches of force (max 60)
Spoke 2: 48.6 x (85 / 90) = 45.9
Spoke 3: 39 x (50 / 90) = 21.6
Spoke 4: 6 x (7 / 90) = .5
Total Side inches of force = 83.1 (maximum 240)

Go-Against-Work Side

Spoke 5: 9" x (42 / 90) = 4.2 inches of force (max 60)
Spoke 6: 30.6 x (57 / 90) = 19.4
Spoke 7: 39 x (80 / 90) = 34.6
Spoke 8: 21 x (32 / 90) = 7.5
Total Side inches of force = 65.7 (maximum 240)

I'm cheating not to change the numbers but only for Tubes 1 and 5, yet those numbers can't change much over 5 degrees of turn. Look at the massive difference in percentage, from 9-percent with precision-vertical spokes to: 83.1 / 65.7 = 1.265 = 26.5-percent larger. I won't include the numbers for this 5-degree turn when calculating the averages. It's here out of interest in knowing the mechanics of, as the wheel turns.

Next:

Do-Work Side, 11.25-DEGREE SPOKES

Spoke 1: 38.5" x (54 degrees / 90 degrees) = 23.1 inches of force (max 60)
Spoke 2: 51.4 x (84/ 90) = 48
Spoke 3: 31.5 x (40 / 90) = 14
Total Side inches of force: 85.1 (max 180)

Go-Against-Work Side

Spoke 4: 3 x (3 / 90) = .1 inches of force
Spoke 5: 12.8" x (22 / 90) = 3.1
Spoke 6: 33.4 x (66 / 90) = 24.5
Spoke 7: 36.6 x (70 / 90) = 28.5
Spoke 8: 12.8 x (22 / 90) = 3.1
Total Side inches of force = 59.3 (max 300)

I'm shocked at how high the percentage is here: 85.1 / 59.3 = 1.435 = 43.5 percent, a whopper of a surplus.

Next, 22.5 degrees (see drawing) from the vertical-spoke position, which is 1/16th the total way around the wheel; therefore, the vertical spoke is now at 22.5 degrees:

Do-Work Side, 22.5-DEGREE SPOKES

Spoke 1: 44.5" x (63 degrees / 90 degrees) = 31.2 inches of force (max 60)
Spoke 2: 49 x (72 / 90) = 39.2
Spoke 3: 25.8 x (27 / 90) 7.7
Total Side inches of force: 78.1 (max 180)

Go-Against-Work Side

Spoke 4: 12 x (13 / 90) = 1.7 inches of force
Spoke 5: 21.6" x (31 / 90) = 7.4
Spoke 6: 37.2 x (77 / 90) = 31.8
Spoke 7: 33 x (57 / 90) = 20.9
Spoke 8: 7.2 x (10 / 90) = .8
Total Side inches of force = 62.6 (max 300)

The good-guy side wins again in the gravity-bite department, this time by 78.1 / 62.6 = 1.25 = 25 percent. After the spokes pass 22.5 degrees, they will return to a vertical-spoke position in one more 22.5-degree spin. And then we have one final set for a spoke position half of a 22.5-degree turn after the 22.5-degree spokes, and this next set is just 11.25 degrees before perfect vertical again at the top. By this time, Tube 8 is on the do-work side; Tube 4 still on the anti-work side:
This next set gives the good-guy side an advantage of only 85.6 - 72.3 = 13.3 inches, and 85.6 / 72.3 = 1.184 = 18.4 percent larger.

Do-Work Side, 33.75-DEGREE SPOKES

Spoke 1: 47.5" x (80 degrees / 90 degrees) = 42.2 inches of force (max 60)
Spoke 2: 43 x (58 / 90) = 27.7
Spoke 3: 10.9 x (13 / 90) 15.7
Spoke 8: 2 x (4 / 90) = .1
Total Side inches of force: 85.6 (max 240)

Go-Against-Work Side

Spoke 4: 23" x (29 / 90) = 7.4 inches of force
Spoke 5: 28.2 x (47 / 90) = 14.7
Spoke 6: 38.5 x (88 / 90) = 37.6
Spoke 7: 25.7 x (44 / 90) = 12.6
Total Side inches of force = 72.3 (max 240)

I've also done the calculations for when the wheel is slightly before vertical-spoke position, so that this starts a full 1/8th turn:

Do-Work Side, 11.25 PRIOR TO VERTICAL SPOKES

Spoke 1: 0 x (0 / 90) = .0
(max 60)
Spoke 2: 49 x (74 / 90) = 40.3
Spoke 3: 46.2 x (62 / 90) 31.8
Spoke 4: 15.4 x (18 / 90) = 3.1
Total Side inches of force: 75.2 (max 240)

Go-Against-Work Side

Spoke 5: 25" x (28 degrees / 90 degrees) = 7.8
Spoke 6: 24.7 x (44 / 90) = 12
Spoke 7: 38.5 x (88 / 90) = 37.6
Spoke 8: 27" x (43 / 90) = 12.9
Total Side inches of force = 70.3 (max 240)

Total Wheel advantage: 75.2 - 70.3 = 4.9 inches

To see how to measure distances and angles lines for different wheel positions by not making new drawings:
http://www.tribwatch.com/photos/perpNewAxleLine.png

### Watta We Got?

Here are all the sets together:

11.25 DEGREE BEFORE VERTICAL: 75.2 - 70.3 = 4.9 angle-modified inch advantage
75.2 / 70.3 = 1.07 = 7 percent

VERTICAL SPOKE 0 DEGREE: 73.6 - 67.1 = 6.5 inches
73.6 / 67.3 = 10 percent

About-5 DEGREE TURN past vertical: 83.1 - 65.7 = 17.4 inches of advantage (don't include this in inch tally)
83.1 / 65.7 = 1.265 = 26.5 percent (don't include in tally)

11.25 DEGREE TURN: 85.1 - 59.3 = 25.8 inches
85.1 / 59.3 = 1.435 = 43.5 percent

22.5 DEGREE TURN: 78.1 - 62.6 = 15.5 inches
78.1 / 62.6 = 1.25 = 25 percent

33.75 DEGREE TURN: 85.6 - 72.3 = 13.3 inches
85.6 / 72.3 = 18.4 percent

45 DEGREE TURN (back to vertical = 0 DEGREE):
Repeat all the above perpetually.

Average Percent Advantage = 89.3 / 5 = 17.9

Let me explain this. As we see it in these numbers, the position at 22.5 degrees before vertical is exactly the same as at 22.5 degrees after vertical. The only thing that changes is the numbering of the spokes; at 22.5 degrees before vertical, Spoke 1 at the top becomes the new Spoke 1 at 22.5 after vertical. So, it means that, 22.5 before vertical has the same advantage as 22.5 degree after vertical, both at 25 percent. But with these numbers, the force drops hard from 25 to 7 percent in just 11.25 degrees of turn, then goes back up to 43.5 percent at 11.25 degrees after vertical. That's a bumpy ride, which has me asking whether I should be measuring distances from the axle rather than the axle line. I should re-do all of these sets in the next update to see how axle measurements look. Even if the numbers are wrong below, the methods by which we can arrive to get the wattage seems sound. My measurements are imperfect, but they are close.

[Insert -- one good reason for rejecting the sets above is that the position at 11.25 degrees before vertical should have the same total-side inches and percentage, yet they do not. As I write here, I've re-done the sets measuring to the axle, and indeed those two positions have the same numbers. I will leave these wrong sets here for anyone to inspect and verify, or make up their own mind, as to whether we should measure to the axle line or axle. I'll also leave this section, though the wattage obtained should be wrong. I'll re-do the wattage calculations after doing the axle measurements.

Also, I should have divided by 4 at the percentage average above, not 5, but was deceived because the percentage was not the same for 11.25 before vertical and 33.75 after vertical. If they were the same, as they should be, I would have realized that we divide by 4 positions, not 5. I'll come back and fix things as I have time. End insert]

Once the spokes are again vertical after 1/8th turn, even though they are different spokes, the percentage numbers repeat themselves, perpetually. We don't need any more calculations. We have the number of surplus inches on the do-work side. That's what I'll use going forward. There was a 66 surplus in inches for 5 equa-distant positions (not including 5-degree turn), wherefore the average inches for use in going forward are: 66 / 5 = 13.2. But where on the clock do we put the average batch at 13.2 inches from the axle?

The lowest advantage point is where the water transfer begins. The numbers don't decline fast due to the partial water transfer, but much due to Spoke 6 coming up big to the 9 o'clock line, during water transfer, beside it's big-gun partner, Spoke 7. Meanwhile, the do-work big gun at Spoke 2 is moving down and away from 3 o'clock, and there are no other big guns until water transfer has completed.

I suggest that the 13.2-inch surplus is to be out at 45-degrees from the axle because that's the average position of an entire side, where its 12 o'clock position is on the 0-degree line while its 3 o'clock position is on the 90-degree line. We can use either 1:30 or 4:30 for this; just fix a batch of water on one of those lines, 13.2 inches from the axle. And how much water do we place there? The whole tube, right? Nope. The whole do-work side. The other 4 tubes, on the anti-work side, have all been placed at the axle with 0 inches in spoke length.

On the 5-degree drawing, there are 4 tubes between 1:30 and 6 o'clock. On the drawing with 45-degree spokes, there are 3 tubes between 1:30 and 6 o'clock, and so there's an average of 3.5 tubes between those times representing 2/3rds of the do-work side. We need to place it ALL as an average, on an average line between lines to 1:30 and 6:00; that average is to 3:45 o'clock. The whole batch goes 13.2 inches from the axle on a line to 3:45. We need to convert the batch's pound-force to the 3 o'/clock line.

When we converted the true inches to angle-modified inches, and then subtracted the anti-work-side's inches from the do-work-side's inches, we moved the WHOLE tube's weight closer to the axle. We don't change the tubes' weight when changing the inches. Half of each tube is filled with 10.9 pounds of water, and so we're dealing with 10.9 x 4 = 43.6 pounds as the whole batch.

Any specific weight at 13.2 inches on the 4:30 line has the same wheel-turning force as when it's at 6.6 inches (half 13.2) on the 3 o'clock line. We don't really care about inches, we want the pound-force. But as the batch is on the 3:45 line, not at 4:30, and as 3:45 is half way between 4:30 and 3 o'clock, we need to reduce the 13.2 inches by half of 6.6 so that it ends up as 9.9 inches from the axle, on the 3 o'clock line.

As 9.9 inches are .825 of 12 inches, we need to decrease the poundage of the batch to 10.9 x 3.5 x .825 = 31.5 pounds. The reason I'm using 3.5 instead of 4 tubes was shown above, and it has to do with the lack of water between 12:00 and 1:30, where water is on the anti-work side while the tubes cross paths with those clock locations. The conclusion is that there's 31.5 pounds, 12 inches from the axle, equivalent in force to 31.5 / (12 / 60) = 6.3 pounds 60 inches from the axle.

This calculator (choose "lbf-ft" for the bottom box) will show that 31.5 pound-force per foot is the work done whether with 6.3 pounds at 60 inches from the axle, or 31.5 pounds 12 inches from the axle. Both get 42.7 joules (choose "J/rad" at the option arrow).

The pounds-force become "foot-pounds" when they travel 12 inches (1 foot for any foreigners reading) around a wheel, 12 inches from the axle. Each pound-force (lbf) produces 1.356 joule while 1 joule is 1 watt per second, and so we have 31.5 x 1.356 = 42.7 watts per second if the 31.5 pounds travels 1 foot around the wheel while being 1 foot from the axle...which is equivalent to 5 feet of travel around the wheel when a weight is 5 feet from the axle. That's why the calculator says 42.7 joules (= watts per second) per radian, for a radian is 1 foot on a 2-foot wheel, or 5 feet on a 10-foot wheel.

Quote: "Conversely one foot pound-force (ft · lbf) is the moment about an axis that applies one pound-force at a radius of one foot." What's a moment? "A moment (also sometimes called a torque) is defined as the 'tendency of a force to rotate a body'. Where forces cause linear accelerations, moments cause angular [pertaining to wheels] accelerations. In this way moments, can be thought of as twisting forces." The water in the tubes "twist" the spokes and tell the axle to make watts. The Water-Watt machine.

Okay, so if a foot-pound is 1 pound turning a wheel from the end of a 1-foot spoke connected to an axle, then 6.3 pounds, 5 feet from the axle, is the force of 5 times more, = 31.5 foot-pounds, because it's 5 times easier to turn the wheel from that far out. But I don't want to know how easy it is for Mr. Pound to turn the wheel. I want to know how many watts Mr. Pound will GIVE to me. If he doesn't tell me, I'm going to punch him out, because I've done a lot of tedious, mind-breaking, maddening work for getting to this point.

When Mr. Foot Pound moves the spoke end 1 foot in one second (when its 1 foot from the axle), he's either burning up, or producing, 1.36 watts. Which one is it? Both, I assume. If gravity pulls with 1 watt, then the water must carry one watt. But what if this is wrong? How can I go forward not knowing? Why, after a month, have I not come across one online example of what pertains to this discussion?

With the fuel or muscle of 1.36 watt, Mr. Pound can move a 5-foot spoke, 5 feet from the axle, 5 feet around the wheel. But what is this possibly saying, that one pound at 5 feet from the axle produces 5 times less watts per foot of travel than 1 pound at 1 foot from the axle?

If we stick a piece of tape 1 foot from the axle on a 5-foot spoke, we will find that it turns 1 foot after the tip of the spoke moves 5. The whole spoke turns, and the tape is still directly between the spoke and the axle after the spoke turns any distance. The tape cheats, keeping up with the tip of the spoke while travelling five times less distance. Which of the two produces more watts?

It seems to me that, while it's 5 times easier to turn the spoke from 5 feet out, one pound on a perpetual-motion wheel should be able to produce 5 times more work in watts from 5 feet out. As 1 pound is stronger for turning the spoke when it's 5 feet from the axle, it should produce 5 times the work. That's half the reason I'm discussing a 10-foot wheel instead of a 5-foot wheel.

The operators of the electrical plant are not going to use a small wheel to produce electricity just because it's all the same, no matter how big or small.

The calculator is telling us that, with 12 inches in the top box for wheel radius, and 31.5 pound-force in the middle box, we can use up or get (which?) 42.7 joules of power per radian. Then, the joules per radian are cut in half when we use 30 inches for the wheel radius (while keeping the 34.7 the same). Go ahead and put the 6.3 pounds above into the calculator with 60 inches at the top, and the joules remains at 42.7. It's magic: 1 pound at 60 inches produces 5 times more watts than 1 pound at 12 inches. It's like energy creation out of thin air.

The 10-foot wheel is like a wheel that has 6.3 pounds of force perpetually at the wheel rim. But it needs to be viewed perpetually at 3 o'clock because this pound force becomes less poundage when not falling straight down to gravity. This pound force is not produced by someone's hand turning the Wheel of Fortune. This wheel is turned by the invisible power of gravity. You can just look at it, and it will turn.

A radian is defined as a distance around the wheel rim equal to the radius. When a 10-foot wheel spins a radian, 60 inches around its rim, it uses or provides (which?) twice as much power as when a 5-foot wheel spins a radian, 30 inches around its rim. Both wheels have the same number of radians per 1 revolution, meaning: the larger the wheel, the more power per 1 revolution. The 10-foot wheel (31.4-foot circumference) uses up and/or produces 47 x (31.4 / 5) = 295 watts per revolution when it has 6.3 pounds connected at 60 inches for gravity to grip/bite.

We can spin a wheel faster than a smaller wheel with less energy input. We've created energy. The calculator is telling us that this happens, but nobody among the experts seems to want to tell this little thing. I've never seen it online.

WAVING FLAG: with the same weight at the rim, there are the same watts per foot of travel between any size wheel, but twice the watts are produced per revolution with the wheel twice the diameter. When the battery load kicks back to slow the wheel, the 10-footer has more power to overcome the kickback. It fills batteries twice as fast, or fills two batteries versus the one. AND MOREOVER, the 5-foot wheel can only harness half the weight in the tubes (unless the tubes are made larger). It's a FOUR-FOLD advantage. But if you need a smaller wheel, instead of building four 5-footers to match one 10-footer, you can do two 7.5 footers in a place having 8-foot ceilings.

If we attach a piece of tape 30 inches from the axle on a 60-inch spoke, that tape makes 1 revolution when the end of the spoke does, but the tape has travelled half the distance over the same period of time, and since wattage produced is identical, by the foot of travel, the larger wheel produced 1 watt twice as fast. Bottom line: BIG WHEEL TWICE AS FAST MAKES 1 WATT TWICE AS FAST even though it's from the same energy input. The only thing we need to do to get it is to build a bigger wheel and load it with more weight.

This wheel may not be deemed worth the money and time for the little we save on electricity from the power grid, but this wheel is cheaper than solar panels. I'll show you how soon.

So is 42.7 watts per second good? Is that 42.7 watts for every second of the day? No. It's per second when the wheel speed is 1 radian per second, either 1 foot of travel on a 2-foot wheel, or 5 feet on a 10-foot wheel. How far will this wheel travel over 1 second? Exactly ?. BUMMER, we don't know.

I'd like to warn that we can't find the watts using the formula for finding kinetic energy on joules. The problem with using this method is that its formula is: .5 x kilograms x meters-per-second squared = joules. Whenever we see seconds squared, it's good only for free-fall of a weight, or the part of a wheel's acceleration, but the water-watt wheel will be at a terminal = stable velocity. Free-fall has acceleration, the wheel doesn't. Free-fall has no wheel torque.

Someone writes: "To calculate wheel torque, you will need to use the formula: Torque = Weight x Wheel Size x Acceleration." What about if there's no acceleration? For the weight factor, it says it can be for "the weight of an object that is being moved by the wheel." I want it the other way around. This formula doesn't seem to apply to a perpetual-motion wheel.

Someone writes: "To calculate inertia for a wheel, you need to know the mass of the wheel and its distance from the axis of rotation." The heavier the wheel, the harder to get it rolling with any specific force, but for a perpetual-motion wheel, the more weight, the better, because it's the weight that turns it.

The perpetual wheel self-limits its own velocity due to the time it takes to transfer its water from tube-end to tube-end. The wheel can't turn faster than the water falls from tube-end to tube-end. This is not wasted power because weight is the basis of total power, not wheel speed. Wheel speeds adds momentum that I see as stored power, not additional power. I see any wheel speed, when charging batteries, as "wasted" power in the sense that it's unused power.

Basically, the lowest speed is when the wheel doesn't have momentum to get past the low-advantage period in each 1/8th of a turn. That speed is called, stopped. The percentage advantages above showed that the wheel has about a 20-percent advantage as soon as water transfer has been complete, then rises to 30-percent at the end of the first 1/16th turn. Then, in the last 1/16th part of the 1/8th turn, the advantage goes steadily down to around 9 percent when the spokes are perfectly vertical with tubes balanced with water (= half transfer of water).

The low power is for brief time between 11.25-degree prior to vertical, and 11.25 after vertical, a total of 1/16th turn. However, the low time is over only about half of that turn, which is 1/32nd of a revolution = 32 / 31.4 = about 1 foot of wheel travel. The good news is, if my math was done correctly, that the wheel still has the advantage on the do-work side through the low periods. The bad news is that the wheel can stop if the battery load and friction are equal in power to the wheel power during the low-advantage periods, that never stop coming.

This is why having two wheels on one axle, with spokes staggered, could be valuable in that it reduces the low time to half of 1 foot of travel. The wheel's momentum, what little it has, will be better able to get past the period when the battery kickback almost equals the wheel's drive power. I don't know how great the benefit is in this regard, for having two wheels, but what we do know: two wheels is better than 1 for total power.

By the looks of things, this wheel will make people change their minds on buying solar panels, especially as this wheel is much cheaper when one builds it. It's going to be cheaper to buy even from the retail store. Nobody can stop people from using this wheel. IF IT WORKS as the numbers suggest here. Nobody can disallow the sale of wood, plastic tubes, and bearings for the sake of saving solar panels from less marketability than they have at present.

The calculated 42.7 watts was for the average advantage, and the calculations above found the average percentage as 17.9. As the 42.7 watts -- lets call it 43 -- was the average within this 1/8th turn, we can find the wattage at the lowest advantage of 7 percent using 7 / 17.9 = .39 of 43 watts, about 17 watts. So, if the battery kickback reaches 17 watts, it's going to make it risky as to whether the wheel will get over the low-advantage hump on momentum alone. Gravity can't come to the rescue until water transfer begins to take place.

I've got to say, the mechanics here are new turf for me, I really haven't got any good grip on what I'm talking about. I just hope I'm staying true. If the battery kickback is 17 watts, the wheel needs to be doing more than 17 watts to keep rolling.

I suppose I have no choice but to build a wheel and hook it up to a battery to discover what it can do.

By the way, if this is helpful for you, from Wikipedia:

The foot-pound force (symbol: ft⋅lbf, ft⋅lbf, or ft⋅lb) is a unit of work or energy in the engineering and gravitational systems in United States customary and imperial units of measure. It is the energy transferred upon applying a force of one pound-force (lbf) through a linear displacement of one foot. The corresponding SI unit is the joule [= 1 watt per second] though in terms of energy, one joule is not equal to one foot-pound.

One so-called "foot-pound per second" is equal to 1.356 watts, but a foot-pound is not a so-called "pound-force per foot". You need to get the difference into your head better than it's in my head. A pound-force per foot is a pound situated 1 foot from an axle. And if it travels 1 foot around the wheel's circle, it becomes a foot-pound. I think I have that straight and right. You'll need to verify any deducing I do here because I'm unable to get more than bare bones and fog on these matters, thanks to google becoming such a mediocre tool for when it comes to delivering requested information.

A foot-pound per second is raising or moving one pound a distance of 1 foot over one second. But does this mean that a 1 pound on the wheel, one foot from the axle, and moving 1 foot around the wheel rim, in one second, produces just 1.36 watts? That's what I'm going with, through the green light. I hope I'm not creating an accident.

The webpage below calculates foot-pounds from "grains." "A grain is a unit of measurement of mass...equal to exactly 64.79891 milligrams." So, I put 64,800 grains into the calculator, which is 1 gram, so little weight, and then I put 300 feet per minute into the calculator too, which is 5 feet per second, but the calculator tells me this makes for 12,947 foot-pounds of energy. Why don't I understand this?

However, a page related to the one above tells: "A grain is a unit of weight equal to .065 grams." That's only 15.38 times less than a gram. So, this time I put 15.38 x 453.6 grams (= 1 pound) x 6.3 pounds = 43,964 grains into the calculator with 300 feet/minute, and the calculator tells me I've got 8,784 foot-pounds, which is ridiculous. Just when I thought I had found an appropriate calculator, I get this nonsense.

Thinking that maybe some person in charge of the page mixed up grains for grams, I put 2,858 into the box because it's 6.3 pounds in grams, and this time the calculator tells that it's equal to 571 foot-pounds. Did I do the calculation wrong that gets 43 watts? 571 foot pounds is 571 x 1.356 watts = 774 watts. But that sounds like way too much for only five feet of travel, at 3.4 mph, on this wheel.

And when we go to the calculator below, we are told the 2,858 grams falling at 5 feet per second gets only 3.3 watts per second. Something seems screwed.

Back to our 43 watts per second per radian, which was found in a calculator as torque on a wheel (using the same 6.3 pounds = 2,858 grams). This calculator gives the formula used by the calculator: 0.5 meter radius x 120 newtons of force x sin(90°) = 60 N·m. Therefore, "sin(90-degrees)" = 1. The use of this 90 degrees, I feel sure, means that the force acts in the direction of the wheel's circle, same as a weight pulled by gravity on the 3 o'clock line. The formula for a perpetual-motion wheel can therefore be reduced to, radius in meters x weight in newtons = newton-meters. Sixty newton-meters are exactly 60 joules.

Therefore, we can do: 1.52 meters (5 feet) x 28. newtons (= 6.3 pounds) = 43 newton-meters = joules = 43 watts per second. As there's 31.4 feet all around that wheel, it's 43 x (31.4 / 5) = 270 watts per revolution. We are now down to the same-old same-old problem: how many revolutions per second will this wheel do? It depends on the battery kick-back from however many volts its kicking back with toward the wheel. Therefore, unless a little birdie from God whispers something into my ear, it's impossible for me to say how many watts this wheel will deliver per second.

But we can dream up a wheel velocity. We take the 270 watts per revolution, and dream a speed of 8 seconds per revolution, then divide the day by 8-second increments, which is 10,800 increments. The math: 10,800 x 270 = 2.9 million watts daily. It's sounds like we just got rich off of gravity and water.

What exactly is 2.9 million watts? Or, a better question, exactly how teensy-weensy is 1 watt? If it were as big as a penny, 2.9 million pennies every day for doing nothing but building the machine and looking at it. That's 121,500 watts per hour, not the same thing at all as 121,500 watt-hours.

We can covert the 43 watts per 5 feet to 8.6 watts per running foot. And so how many do we have? It depends on how many feet per second the wheel travels. If we can't know that speed, we can't know the running watts, nor the watt-hours.

How many watts will the wheel be pumping into the battery when its spinning 1 revolution per 8 seconds? I suppose that's a question of efficiency in converting mechanical power to DC watts, and a question also for a generator expert. But at least we have some watt numbers to go on with some arbitrary speeds.

If the wheel could do 5 feet per second, that would be 43 watts per second, which might be the definition of 43 running watts. If it is, then one gets 43 watt-hours every hour. That speed is 1 revolution per 6.28 seconds, and 16 RPM, which I think is do-able for this wheel. I soon show what a big deal these little 43 pennies could be when mixed with some lead.

Alas, this wheel can't do 60 RPM because that's 31.4 feet per second, roughly the speed of straight-down falling water AFTER a full second of fall, yet the water in the tubes can't fall that fast because anything falling to gravity travels much slower than 32 feet per second in the first tenth-second of fall. After looking at this debilitation of the perpetual-motion wheel, I figured that a speed of 1 revolution per six seconds is likely do-able for a 10-foot wheel. I didn't know this a couple of weeks ago.

I had done some math to show that we can't use the kinetic-energy formula for our needs here. A one-second spin of this wheel over 5 feet converts to 1.52 meters per second. The kinetic-energy formula for this velocity is: .5 x 2.9 kg x 1.52 x 1.52 = 3.35 watts per second, where 2.9 kilograms is the 6.3 pounds of water (found above) at 3 o'clock's rim moving 1.52 meters per second. In other words, as the 3.35 watts per second does not match the 43 watts per second given by the calculator for 5 feet of travel, it verifies that we cannot use the kinetic-energy formula.

We saw a calculator above give 3.3 watts too, and the one at this page confirms that the basis of this result is by using the kinetic-energy formula: K.E. = 1/2 m v2.

Perhaps what this formula is achieving is to show how many watts 6.3 pounds produces when free-falling at 1.52 meters per second, which is the speed after a small fraction of 1 second after it's dropped. For, the speed of any falling object is 4.9 meters per second at the end of the first second of fall. We don't want this picture. We want the second-by-second, or minute-by-minute, watts of a non-accelerating poundage. The kinetic formula is good for water falling on a water wheel, but then it needs to be incorporated with the radius of the wheel.

The calculator below tells that if 2.9 kilograms (weight of 2.9 liters of water) of water falls 1.52 meters, it produces 43.2 watts. That's almost exactly the same watts as I have for the 2.9 kilograms turning (mainly downward, straight-down fall) 1.52 meters on the 10-foot wheel. Drats, I was hoping it was wrong to have only 43 watts.

No matter how long the water takes to transfer, it's energy of fall probably goes into turning the wheel when it hits the bottom of the tube, and moreover the tube is attracted by gravity throughout the transfer.

Question: if the wheel spins at 5 feet per second, does it mean necessarily that it's pumping 43 watts into the generator? Or, when this calculator (or when the foot-pound calculation) gives 43 watts per second over 5 feet, is this the speed and wattage of the wheel with or without a load? With a load, I suppose. I've got to assume that, while spinning without a battery load, the wheel, if not debilitated by water-transfer restriction, will be at terminal velocity when friction sends a kickback to the wheel of 43 watts per second.

Therefore, it seems feasible that, when the wheel is doing half that speed, at 2.5 feet per second, while on a battery load, it's loosing half its wattage to the battery. The formula is: 43 total watts x ((5 feet total - 2.5 lost) / 5 feet)) = 21.5 watts.

Then, when travelling at one foot per second while on a load, the wheel is losing: 43 total watts x ((5 feet total - 1 lost) / 5 feet)) = 34.4 watts to the battery. The formula works. but will the wheel still be turning when sending this 4/5ths of its power to the batteries? It will have about 9 AVERAGE watts of excess power at that time, with the low periods being about half that. I don't think the wheel can go that slow unless backed up with a second one on the same axle.

Repeat from above: "As the 43 watts was the average within this 1/8th turn, we can find the lowest advantage as about: 9 percent / 20.5 percent = .44 of 43 watts, and that's roughly 19 watts. So, if the battery kickback reaches 19 watts, it's going to make it risky as to whether the wheel will get over the low-advantage hump on momentum alone." The battery kickback will be about 15 volts, because the battery charger charges at that voltage, but volts and watts are not the same thing.

I have no idea how the charger converts the wheel's wattage to voltage, or how many of the wheel's watts is needed to form 15 volts. I think we need to ask the generator about that. We need to know how much wheel RPM we will convert to generator RPM for optimum performance.

The 10-foot wheel has double the ability to overcome total wheel stoppage at the slowest speeds, as compared to the 5-foot wheel. That is, if both wheels have the same battery load coming against them, the 10-foot wheel has double the speed/momentum at that time.

We can see why this wheel needs lots of weight, but more importantly why it needs plenty of spokes / tubes. The more spokes / tubes it has, the higher the lowest-advantage level. There is a difference between a wheel with 8 spokes and 12 tubes versus 12 spokes and 12 tubes. The latter is better, I tentatively think, decreasing the time between lowest- and highest-advantage points. If the wheel can't get over the low-advantage hump for lack of spokes, having a million pounds of water in the tubes won't make it spin once stopped.

It would be easy to put two wheels on the same axle rod; just stagger the tubes from one wheel to the other so that it's equivalent to one wheel with 16 spokes and 16 tubes. This reduces the 1 foot of low-period turn to half as much, and doubles the momentum power for getting across the low-period bridge.

If two of these wheels can charge batteries well, that's a lot of power, but one should stick to one wheel at first, and replace about a quarter of the water with lead balls. Lead weighs 11 times more than water so that the sacrifice of a quarter of the water, weighing 10.9 / 4 = 2.7 pounds, is replaced with 2.7 x 11 = 30 pounds. Have we got your attention? If the low-period bridge is an issue at all (I don't know whether it is), lead just allowed us to cross it. But is lead mixed with water noisy in plastic tubes? Lead resists corrosion, perfect.

The water and lead combo won't fall in the tube as much as slide over, which is quieter than falling. The faster the wheel, the more the liquid falls rather than slides, yet even at 3 o'clock, the tube is at a 45-degree. The water starts to transfer just before 1:30 on this wheel design. The lead or steel (8 times heavier than water) will be the last to slide-fall, into the liquid already down the tube. Sound-proof a shed wall between the wheel and the house, with wheel in the shed.

GOOD NEWS: I've just read that even a solar panel producing 5 measly watts maximum can charge a 12-volt battery (if not mistaken, the battery charger between solar panels and batteries is a "generator"). The bad news is, I don't know whether the wheel can get the volts up to the needed 15 when it's turning slow. This is a question for the small windmill to answer.

Somebody at Todays' Homeowner writes: "Wind turbines generally make between 10 and 20 revolutions per minute." That's on a load, yet, if even a 5-watt panel can charge a 12-volt battery, the battery kickback is very minor indeed. I didn't mean to scare you with "kickback." It doesn't look scary at all, suddenly. At 5 feet per second on this wheel, it's 9.5 RPM, and even though it can possibly do better, it's my predicted speed when NOT on a load. How much will the batteries slow it down?

### Twenty-Four/Seven

If someone has eight solar panels each putting out a maximum 100 watts (total 800 watts), it's enough to run water pump, lights, and a small fridge and freezer at least most of the day...but twice as much in southern latitudes. There are average regions that get about two "peak sun hours" daily, on average, meaning the equivalent to two hours of sun directly over the panels and shining on them at a dead-on angle (not usually the case when they are on a roof). This means that, on average, regions with two sun hours can get 800 x 2 = 1,600 watt-hours daily with those eight panels.

I don't know how many of its tentative 43 watts this wheel can use in charging before it stops turning, but I'll use the whole 43 for the math to give some indication of the potential: 43 watt-hours x 24 hours = 1,030 watt-hours daily, getting close to the 1,600 from the 8 panels. Can you see how that can make northerners happy in a crisis situation? You don't need to monitor cloud conditions days in advance, because water-watt wheel keeps spinning rain or shine. Twin wheels on one shaft could out-do the 8 panels.

If you have just 11 pounds of lead balls per tube, you can quadruple the tube's weight, and therefore quadruple the wattage. Here's a company selling 100 pounds of lead "granules" for \$350 (world is full of thieves). This wheel has the most attraction in northern latitudes, all of canada, and much of Europe and Russia.

You can get 1.64 times as much water weight by using 5-inch tubing instead of 4. That's 64-percent more power. Dare you try 6-inch tubes? Shouldn't 4 x 4 wood spokes handle them? Sure. You're the boss, you call the shots.

As this wheel is a security measure, it's justified to spend more than you'll get pay-back by the amount of electrical production. Mixing mercury (very expensive) with water is like mixing steel with feathers, because water is 13.6 lighter than mercury. For the cost difference, lead is the way to go even though mercury may be quieter. Mercury mixes with water without issue.

800-1000 watts of solar panels can feed 8 large deep-cycle batteries even in canada. I'm now getting the impression that this wheel can feed the same eight batteries, wow, if only the battery kickback doesn't slow it much. I'm going to need to get very familiar with slow / small windmills to get a better feel on this slowness issue.

Ahh, this page on small windmills has this graph showing that, best as I can tell, the windmill starts producing power at something like 15 watts. The page says that a speed of 3.6 meters/sec (11.8 feet per second), the small windmill he's talking about, gets 50 watts. The calculator says I have 43 watts per second per 5 feet of wheel travel. Why the difference? It all depends on the force on the wheel, and he doesn't say what force he has on it aside from wind speed. And usually, the wind force is done in psi, not foot-pounds. To compare with the water-watt wheel, I've been trying to find how many pound-force per foot windmills take to produce so-many watts. No luck yet.

This wheel's pound-force is capable of getting more than 9.5 RPM if the water-transfer debilitation didn't affect it. The debilitation does not remove any power from the wheel. It's always there, but restricted in the background, only a fraction of it gets out...until the battery load slows it from its terminal velocity. Once the load slows it, the wheel starts to use ALL of its camel power. I don't think it's correct to jump to conclusions in thinking that the slow-down at terminal velocity is a gobbler / destroyer of power.

The debilitation may actually be an excellent feature in that it holds back the full power. Charging batteries with lower power per unit time may be beneficial in multiple ways. Less wastage comes to mind. Less wear and tear on the wheel. Less noise from the wheel. Smaller wiring needed. Smaller battery charger needed. But when you want power faster, sorry, the wheel has a debilitation. The point is, the slow-down debilitation should not be a concern.

The water-watt wheel cannot do much better than 9.5 RPM; it needs a pulley system to get the generator to 2,000 RPM or better. Let's call it 10 RPM. We can use a 12-inch (or 15") pulley connected (with belt) to a 1-inch pulley that is itself on a turning shaft (with bearings) along with a second 12-inch (or 15) pulley. The latter is connected to a second 1-inch pulley on the generator shaft. The wheel's 10 RPM turns into 120 RPM (or 150) at the first 1-inch pulley, and then to 1,440 RPM (or 2,250) at the generator shaft. So long as the wheel has power to put voltage into the batteries when turning the generator at 1,440 (or 2,2350) RPM, we're good.

The 43 joules per radian that was provided by the calculator can also give the result in newton-meters, and, I've just learned: 43 joules per radian (per 5 feet of travel) is exactly 43 newton-meters. This page, with a small windmill rated at 300 watts, says that the "Start Torque" is .1 newton-meter. I assume the start torque is the cut-in point where it starts to generate power. Into a battery, right? The unit comes with its own gearing system to get the RPM way up from the windmill RPM.

Instead of 1,600 watts daily, a video below has a water wheel producing about 1,600 watts constant with about six water buckets holding what, at first glance, looks like 3 gallons each as water passes over. Those 18 gallons would amount to 25 pounds perpetual per bucket (if 6 of them hold 3 gallons), and 150 pounds total, but I don't think he has 3 gallons per bucket. His 6 buckets are from 12:30 to 4:30 o'clock, roughly, and my wheel's poundage is one unit of 6.3 pounds at 3 o'clock 60 inches from the axle.

He only has one bucket at a time at 3 o'clock. The average gravity bite for 6 buckets from 12:30 to 4:30 is less than half so that his 150 pounds comes down in value to under 75. That would be the number if all his buckets were between 1:30 and 4:30, but as one bucket is at about 12:30, I'm going to ballpark that his 150 pounds acts like 65 when all lumped as a batch at 3 o'clock. The score so far is 65 to 6.3 in his favor.

His wheel looks to be doing about 1 revolution per 3 seconds, and so if mine does 1 revolution per 6 seconds, then the two wheels are equal in force in this regard. He has twice the speed, but I have twice the wheel diameter.

His wheel looks to be about 5 feet round, and so his wheel's pounds are about half of mine, meaning the tentative score is now about 33 to 6.3.

With 75-percent water and .25 lead in all my tubes, they would weigh about 37 pounds each, and as that's 3.4 more than the 10.9 pounds of water they have that was used to calculate 6.3, the latter now goes up to 6.3 x 3.4 = 21.4 pounds. The score is now 33 pounds to 21.4 pounds in his favor, meaning that, if indeed he has 3 gallons per 6 buckets, my wheel could get 1,600 x (21.4 / 33) = 1,038 watts...with the lead. He can't put lead into his wheel, but we can have it 24/7/365 without wastage of one granule.

That 1,038 watts (= 1,038 watt-hours every hour) is making me ask questions. It doesn't look correct unless he has a lot more than 3 gallons per 6 buckets. It doesn't look correct because the lead increases the weight by 3.4 times, wherefore it should increase the 43 watt-hours by the same to 146. The other option is that I went wrong in arriving to 43 watts. But how could I be so far off? I do hope I made a mistake; I do hope it's more than 43 running watts.

I'm thinking of re-doing the math sets using distance from the axle instead of distance from the axle line. I may have that up below, in this update, by Wednesday of this week. I've got to take another good-hard look at everything, please forgive my mistakes thus far.

I can't see how much water is in the buckets due to the splashing, but while his water-feed trough is probably pouring at least 3 gallons per second, it doesn't mean each bucket gets filled with that much. At .3 RPSecond, his wheel, with about 18 buckets all around, is therefore moving 6 buckets per second. If the water source is even 10 gallons per second, he can't have 3 gallons per bucket. The trick is to figure how much water the trough delivers per second, but this is too hard for me to figure with any certainty. The man should have said something because this "little" detail is what viewers would like to know if they have a water-wheel potential at their property.

There's a lot of water being struck and sent forward by and from the wheel when it's going fast in the 16th minute. I'm a little relieved, having watched the video a second time. I don't know if he's getting even 1 gallon per bucket in the 16th minute, judging by watching buckets pass by at the mouth of the trough (or ramp). He may be getting half of his wheel speed from the water velocity as it comes off the ramp. I think he gets more water in the buckets when the wheel is slower.

> If you make the video play at quarter speed, then, like me, you may see that there isn't much water pouring out the buckets at the bottom of the wheel. It doesn't look like a gallon to me out of each bucket. In the 24th minute, with the video at quarter speed, I can't see much water coming out the buckets. You can see, at the top of the wheel, that the part of the buckets intended to hold water are knocking half of it away frontward before it can drop into the buckets. If he's doing 1,500 watts with a gallon only per bucket, I'm very encouraged.

Near the start of the video, he shows that his induction motor-generator produces over 1,700 watts at 730 RPM = 12.2 RPSecond. I gather that the steel box beside his motor, beside the water wheel, is his gear box for getting his .3 RPSecond wheel up to 750-ish RPM.

At the speed needed for over 1,700 watts, he's getting 185 volts. This is a very good chunk of power, and like he says, it's 24/7, 365 days a year, doesn't take vacations or complain about high taxes. It doesn't get scurvy or West Nile virus. He needs to get the ramp higher so that water comes down at an angle to the buckets, filling them more. He's got lots of water so that he could have made the wheel twice as long at the axle, doubling the water weight in the buckets, at the cost only of stronger rod and bearings.

In the 9th and 10th minute, a good shot of the buckets and the size of the wheel.

He's using an induction motor as a generator, though it needs rightly-sized capacitors to ignite the motor's electromagnets. Experts in the field would understand that. We won't need a motor that large. He appears to be scrapping his white generator that's a typical one for windmills; he doesn't tell, in this video anyway, why he's scrapping it for his water-wheel.

### Math Sets Re-Done with Axle Distances

Once I figured that the measurements to the axle line are the wrong way to do things, the sets were re-done with water-to-axle measurements instead. Almost all of the distances will be either 52 inches for the do-work side, and 40 inches from the anti-work side. As 52 / 40 = 1.3 = 30 percent, the maximum per do-work-side tube is 30-percent advantage, a situation that would come about if total figures for angles were equal on both sides. Angle measurements remain the same in these sets as those above.

Do-Work Side, 11.25 PRIOR TO VERTICAL SPOKES

Spoke 1: 52 x (0 / 90) = .0
= 0 inches of force (max 52)
Spoke 2: 52 x (74 / 90) = 42.7
Spoke 3: 52 x (62 / 90) 35.8
Spoke 4: 52 x (18 / 90) = 10.4
Total inches of force: 88.9 (max 208)

Go-Against-Work Side

Spoke 5: 40" x (28 degrees / 90 degrees) = 12.4 (max 40)
Spoke 6: 40 x (44 / 90) = 19.6
Spoke 7: 40 x (88 / 90) = 35.1
Spoke 8: 40 x (43 / 90) = 19.1
Total inches of force = 86.2 (max 160)

Total Wheel advantage: 88.9 - 86.2 = 2.7 inches
Percent advantage: 88.9 / 86.2 = 1.031 = 3 percent (max 30-percent)

Do-Work Side, VERTICAL SPOKES

Spoke 1: 46" x (27-degree / 90-degree) = 13.8 inches of force (max 52)
Spoke 2: 52 x (85 / 90) = 49.2
Spoke 3: 52 x (50 / 90) = 28.9
Spoke 4: 52" x (8 / 90) = 4.6
Total Side inches of force = 92.3 (maximum 208)

Go-Against-Work Side

Spoke 5: 46" x (27 / 90) = 13.8 inches of force (max 40)
Spoke 6: 40 x (57 / 90) = 25.3
Spoke 7: 40 x (80 / 90) = 35.6
Spoke 8: 40 x (32 / 90) = 14.2
Total Side inches of force = 87.7 (maximum 160)

Total Wheel advantage: 92.3 - 87.7 = 4.6 inches
Percent advantage: 92.3 / 87.7 = 1.05 = 5 (max 30-percent)

Do-Work Side, 11.25-DEGREE SPOKES

Spoke 1: 52" x (54 degrees / 90 degrees) = 31.2 inches of force (max 52)
Spoke 2: 52 x (84/ 90) = 48.6
Spoke 3: 52 x (40 / 90) = 23.3
Total Side inches of force: 103 (max 156)

Go-Against-Work Side

Spoke 4: 40 x (3 / 90) = 1.4 inches of force (max 40)
Spoke 5: 40" x (22 / 90) = 9.8
Spoke 6: 40 x (66 / 90) = 29.4
Spoke 7: 40 x (70 / 90) = 31.2
Spoke 8: 40 x (22 / 90) = 9.8
Total Side inches of force = 81.6 (max 200)

Total Wheel advantage: 103 - 81.6 = 21.4 inches
Percent advantage: 103 / 81.6 = 1.26 = 26 percent (max 30)

Do-Work Side, 22.5-DEGREE SPOKES

Spoke 1: 52" x (63 degrees / 90 degrees) = 36.4 inches of force (max 52)
Spoke 2: 52 x (72 / 90) = 41.6
Spoke 3: 52 x (27 / 90) 15.6
Total Side inches of force: 93.6 (max 156)

Go-Against-Work Side

Spoke 4: 40" x (13 / 90) = 5.8 inches of force (max 40)
Spoke 5: 40 x (31 / 90) = 13.8
Spoke 6: 40 x (77 / 90) = 34.2
Spoke 7: 40 x (57 / 90) = 25.4
Spoke 8: 40 x (10 / 90) = 4.4
Total Side inches of force = 83.6 (max 200)

Total Wheel advantage: 93.6 - 83.6 = 10 inches
Total percent advantage: 93.6 / 83.6 = 1.12 = 12 percent 7.8

Do-Work Side, 33.75-DEGREE SPOKES

Spoke 1: 52" x (74 degrees / 90 degrees) = 42.8 inches of force (max 52)
Spoke 2: 52 x (62 / 90) = 35.8
Spoke 3: 52 x (16 / 90) 7.5
Spoke 8: 52 x (2 / 90) = 9.2
Total Side inches of force: 95.3 (max 208)

Go-Against-Work Side

Spoke 4: 40" x (26 / 90) = 11.6 inches of force (max 40)
Spoke 5: 40 x (47 / 90) = 20.9
Spoke 6: 40 x (88 / 90) = 39.1
Spoke 7: 40 x (47 / 90) = 20.9
Total Side inches of force = 92.5 (max 160)

Total Wheel advantage: 95.3 - 92.5 = 2.8 inches
Total percent advantage: 95.3 / 92.5 = 1.03 = 3 percent

It makes perfect sense that the same 3 percentage is obtained with the 11.25-degrees before vertical-spoke position and the 33.75 degrees after vertical-spoke. These positions are 1/8th turn apart, and so the same percentages repeat themselves perpetually every 1/8th turn. There should be the same total numbers no matter where we start and finish a 1/8th turn.

This 3-percent of surplus power for the do-work side is the weakest link on this wheel. I don't like it for the purposes of charging batteries. I would rather see steady power rather than fluctuating as it does. Here's a bird's eye view of things:

11.25 DEGREE BEFORE VERTICAL: 88.9 - 86.2 = 2.7 angle-modified inch advantage
88.9 / 86.2 = 1.027 = 3-percent surplus

VERTICAL SPOKE 0 DEGREE: 92.3 - 87.7 = 4.6 inches
92.3 / 87.7 = 1.05 = 5-percent surplus

Water is balanced at vertical, but almost-immediately afterward transfers to add maximum power to do-work side, and transfers also on the other side to lose power for anti-work side, explaining the sudden, high 26 percentage that comes of it, in the below.

11.25 DEGREE TURN: 103 - 81.6 = 21.4 inches
103 / 81.6 = 1.26 = 26-percent surplus

22.5 DEGREE TURN: 93.6 - 83.6 = 10 inches (don't include in averaging)
93.6 / 83.6 = 1.12 = 12-percent surplus (don't inc. in averaging)

33.75 DEGREE TURN: 95.3 - 92.5 = 2.8 inches
95.3 / 92.5 = 1.03 = 3-percent surplus

1/8th turn completed; after one more 11.25-degree of turning it's back to 45 DEGREE TURN = vertical spokes = 0 DEGREE.

Total 1/8th Turn Advantage = 38.7 inches
Average Percent Advantage = (3 + 5 + 26 + 12) / 4 = 11.5.

This is a better look than the roller-coaster ride when measuring to the axle line. The numbers now suggest a smooth transition from high power to low and back up to high again. Whether the highest point is exactly 26 percent at the 22.5-degree point, I don't know without a couple of more measurements, but I'm all measurement pooped out. Forget it, we can work well-enough with these numbers. It would be nice to know what the lowest number is, and so here's the scoop on 5-degree before vertical:

Do-Work Side, 5-DEGREES BEFORE VERTICAL SPOKES

Spoke 1: 52" x (9 degrees / 90 degrees) = 5 inches of force (max 60)
Spoke 2: 52 x (79 / 90) = 45.6
Spoke 3: 52 x (57 / 90) = 32.9
Spoke 4: 52 x (14 / 90) = 8.1 inches of force
Total inches of force: 91.5

Go-Against-Work Side

Spoke 5: 40" x (29 / 90) = 12.9
Spoke 6: 40 x (53 / 90) = 23.5
Spoke 7: 40 x (83 / 90) = 36.9
Spoke 8: 40 x (37 / 90) = 16.4
Total inches of force = 89.6 (max 160)

Inch advantage: 91.5 - 89.6 = 1.9
Percent advantage: 91.5 / 89.6 = 1.02 = 2-percent surplus

This 2-percent outcome suggests that power doesn't drop below 3 percent by much, but goes up from there to the 5 percent at vertical-spoke position. This is more proof, I think, that wattage figures should be obtained by measurements to the axle, because there is a smooth deviation of power as the wheel rolls on.

However, I really am getting nervous with this extremely-low power for 1/32nd of a turn, which for a 10-foot wheel is 1 foot of travel. The momentum together with this low gravity bite is all there is to keep the wheel spinning against a battery load. I really don't like this. Starting with vertical spokes, and going forward in 1/32-turn increments, the percentage changes from 5 to 26 to 12 to 3, then back to 5 to 26 to 12 to 3, perpetually.

When measuring the 38.7 total inches to the axle line for all 4 positions, the average is 9.7 inches, meaning that I'm not choosing the axle-measurement method due to it being worth more watts; i.e. 9.7 is smaller than the 13.2 obtained by axle-line measurements, and worth less watts. But where on the clock do we put the average weight, as 1 batch, 9.7 inches from the axle?

There is very difficult problem in choosing how much water weight to include in a batch. I don't think we can use all 4 tubes because there is an average of only 3.5 tubes doing all the work on the do-work side, from vertical spoke to vertical spoke. There's more than one way to skin this cat.

At vertical-spoke position, the only 3 tubes doing any work are almost exactly at 45 degrees, 90 degrees, and 45 degrees, a nice situation because both tubes at 45 degree are worth exactly half the force of the 1 tube at 90 degree; the latter is at 3 o'clock. Therefore, this situation is identical to having the water of 2 tubes both at 3 o'clock. We can start with this.

Thus far, we have 2 tubes of weight, 9.7 inches out toward 3 o'clock, but we're not done. The sets above show that, after starting with a 5-percent advantage at the vertical-spoke position, torque increases throughout the 1/8th turn, to the next vertical-spoke position, by an average of: (5 + 26 + 12 + 3) / 4 = 11.5 percent. As that's and increase 5.5 percent, I think we increase the 21.8 pounds of water in 2 tubes to 21.8 x 1.055 = 23 pounds.

When we converted the true inches to angle-modified inches, and then subtracted the anti-work-side's inches from the do-work-side's inches, we moved the WHOLE tube's weight closer to the axle. We don't change the tubes' weight when changing the inches. Half of each tube is filled with 10.9 pounds of water, and so we're dealing with almost 44 pounds of water on the do-work side.

We don't really care about inches, we want the pound-force. As the batch is now working out to 23 pounds at 9.7 inches from the axle, we need to do: 23 x (9.7 / 60) = 18.6 pounds at 12 inches from the axle.

When someone turns the Wheel of Fortune with 18.6 pounds of force with a hand on the wheel's rim, he/she turns it in the direction of the circle. Gravity can't do that unless the water is at the rim at 3 o'clock. That's why I need the whole water as one batch at the 3 o'clock rim. The 18.6 pounds at 12 inches upon the 3 o'clock line is equivalent in force to 18.6 x (12 / 60) = 3.72 pounds at the 3 o'clock rim. It sounds so measly. What are we going to do with that little thing?

This calculator (choose "lbf-ft" for the bottom box) will show that 18.6 pound-force per foot is the work done whether with 3.72 pounds is at 60 inches from the axle, or whether there's 18.6 pounds 12 inches from the axle. Both show 25.2 joules (choose "J/rad" at the option arrow) of force, which is 25.2 watts per second.

The wattage is smaller than the 43 obtained with axle-line measurements. It seems like the only good hope is to add some material into the water. Sand doesn't weigh much more than water, but it is cheap and quiet. One could line the inner tubes with rubber (keep noise down) and go with lead granules. There's got to be cheap-grade lead pellets, anything as long as it works.

If you can get lead for 3 dollars per pound, and you put 33 pounds in the tubes with water, for say, 40 pounds per tube total, and if this gets 100 running watts all day every day, it's a kilowatt-hour every 10 hours. Just to make the math easy: if one is paying 10 cents per kilowatt-hour on the grid, it'll take 1,000 increments of 10 hours, about 42 days, to pay for one tube's worth of lead by the value/savings.

The small challenge is in connecting a 40-pound tube to the spokes. The tubes will connect all spokes at their perimeter, but for 40 pounds per tube, you need a reliable connection method. There's options.

However, we need to take a hard look at the weakest link in this wheel if indeed the advantage goes down as low as 2-5 percent over a foot of wheel travel. This is where two wheels with staggered spokes could be very useful in reducing the one foot to 6 inches, and doubling the momentum. But what size rod and bearings would be needed to support 16 tubes each with 40 pounds?

The rod should be supported between wheels to keep it from bending. And minute amounts of rod bending (applies pressure on bearings) will increase friction. I'm thinking that a 1.75-inch diameter rod, little more than one foot long, with a bearing in the middle, three bearings in all, could support two wheels where each wheel is only 4 inches wide. Here's a rudimentary drawing:
http://www.tribwatch.com/photos/perpWheelBearingMount.png

"Radial ball bearings are designed to withstand forces that are perpendicular to the direction of the shaft, or radial loads. Radial bearings are the most common type." Pillow-block bearings seem ideal. But the world-corporate thieves are jacking everything through the roof like there's no tomorrow. Maybe they feel that tomorrow is just about the end of all things. Don't pay for bearings rated for mega-pounds when you don't need that much.

There's going to be a problem with frost heave in cold climates. What happens when the frost lifts the floor upon which this machine is mounted? Does this mean that one would need a concrete slab too, one that doesn't crack? That's not cheap. I'm realizing that this could be a problem in winter for using a wood-floor deck. If only one corner heaves, it could hurt the bearings. If all four corners heave in differing amounts, it could harm the bearings.

If you are on a river, here's a type of water-wheel pump I've not known before:

I'll leave my email in case you can see where I'm going wrong with the watt quest: johny@xplornet.com.

NEXT UPDATE

Here's all four Gospels wrapped into one story.

For Some Prophetic Proof for Jesus as the Predicted Son of God.
Also, you might like this related video: