Worcester wheel

[harry ravi]

Hello,

Yes it's been a long time since I last gave my news, still no proto because no time.

Thank you for the mechanical study, it seems very complete, I said seems because I do not understand everything.

Question: what is a negative couple? normally a couple at 0 corresponds to a zero force .... a negative couple would mean that the applied force is opposite?
According to my memories of physics, for a system in balance, the sum of the forces must be equal to 0 and not lower or equal to 0?

In addition, I have not seen the concept of time in order to convert this torque into kinetic power.
If ever the announced torque is in Nm-1, it will be necessary to know the power in Wh, even negative.

In addition, it seems to me that the calculations were made from a weight of 1kg and not from a weight of 1000g (the couple this calculation well with grams right?). In this case the torque announced by your excel sheet announces a torque of -128 Nm.

I do not contradict anything, especially on the part of scientists but I wonder about this couple of -0.129 Nm which can be due to the approximation of the calculations (friction forces not included, ....) proving that the machine has a torque equivalent to 0 or the machine has a low torque very close to 0.

In any case, thank you for your interest in this subject.

[harry ravi]

Using your spreadsheet, I get with:
40 weight of 1000g, a square dimension of 0,10m and a complete rotation time of 60s: -38,458 .. Nm of torque (it seems to me that the measurement in radian is done counterclockwise, may be the cause of this negative value)

And with a formula: approx 5 kWh of power

I modified / added / used:

side size of a square 0,1 (cell B4)
mass in grams (cell B6)
rotation time 60s (cell B7)
distance traveled by force (R1 * Pi / 2) + (R2 * Pi / 2) (cell G12)
total torque (cell C16)

power calculation (ABS (C16) * (G12 / B7)) * 3600 in Wh (cell I4)
power kWh I4 / 1000 (cell I5)


http://www.hiboox.fr/go/images/image-perso/worcs,9fa55eaea26f588d21c1fef566857650.jpg.html

Again, tell me if I'm wrong.

[Sadistic Arnaud]

Hi Harri the delighted, it's good to ask questions, it shows at least that you are interested in my work;)

what is a negative couple? normally a couple at 0 corresponds to a zero force .... a negative couple would mean that the applied force is opposite?

Basically, yes.

To say with the hands, a couple is at an angular displacement what a force is at a displacement quite short. I confuse, like many Mechanics, the notion of torque with that of moment. The only difference is that a couple is produced by a system of forces whose sum (vectorially speaking) is zero, which is generally not the case for a moment. A good example of torque is the one you need to produce to screw a spare wheel with a cross: your two arms exert two forces equal in intensity and direction, but opposite in direction.

We calculate the moment of a force with respect to a point by taking the product of the force by the lever arm passing through this point. From a mathematical point of view, this amounts to performing what is called a cross product between the lever arm vector and the force vector concerned. A couple is generally signed: by convention, the sign is denoted plus when the force "turns" around the point in a counterclockwise direction (counterclockwise) and negative otherwise. Often a positive torque is considered as driving torque, and a negative torque as resistant: this is somewhat the case with the Worcester wheel.

According to my memories of physics, for a system in balance, the sum of the forces must be equal to 0 and not lower or equal to 0?

Indeed, my sheet shows that at any angle does not necessarily correspond to a zero torque: according to this angle, we have positive or negative values. If you plot the torque as a function of the angle, you will see that it is a periodic function (with a period of 9 °) and perfectly symmetrical with respect to the x-axis - it is in fact a tooth function of triangular saw. What it means physically: there are a series of particular angles (there are exactly 40) for which the torque is zero. If we position the wheel at one of these angles, it will not rotate: it will be in equilibrium. If, on the other hand, you position the wheel at another angle, then the wheel will start to rotate, then oscillate around the nearest point of equilibrium, until it stabilizes at this point of equilibrium.

FYI, I recently modified this sheet to take into account perfectly square compartments rather than trapezoidal compartments, in order to comply with a directive from Jim Hackenberger. The result is surprising this time: the torque produced by the weight arrangement is zero for any angle. The wheel is therefore always in equilibrium, whatever the angular position imposed on it.

In addition, I have not seen the concept of time in order to convert this torque into kinetic power.
If ever the announced torque is in Nm-1, it will be necessary to know the power in Wh, even negative.

Exactly, the concept of time is not explicitly present, for the reason that I worked in static. If we consider that the wheel behaves as static (so no dynamic effects, which are mainly in the swing of the weights and in the oscillations of the wheel) nothing prevents you from writing that the angle of the wheel theta = RPM * pi / 30 * t where t is the time in seconds. (RPM = revolutions per minute). In this case, the driving power of the wheel as a function of time is equal to the product Torque [theta (t)] * RPM * pi / 30. As it is nothing but the product of the couple by a constant, it means that the power of the wheel which is driving on half of the angular sector of 9 ° becomes very exactly resistant on the other half. Your wheel neither produces nor consumes anything; in fact, it consumes what it produces. This is the best case: in reality, since there is friction on the axle, your wheel will consume more than it will produce, which, of course, was not the desired effect at the start.

In addition, it seems to me that the calculations were made from a weight of 1kg and not from a weight of 1000g (the couple this calculation well with grams right?).

What difference do you make between 1000g and 1kg? Otherwise, a couple is calculated by making the product of a force in newtons by a lever arm in meters, the result being in Nm. The weight in newtons is expressed by the relation P = mg where g = 9.81 m / s ^ 2 and m is the mass in kg.

I hope not to have "taken the lead" too much with all my theoretical-mathematical considerations, and wish you good evening.

[harry ravi]

Nice explanation but I would like to discuss the last paragraph

What difference do you make between 1000g and 1kg? Otherwise, a couple is calculated by making the product of a force in newtons by a lever arm in meters, the result being in Nm. The weight in newtons is expressed by the relation P = mg where g = 9.81 m / s ^ 2 and m is the mass in kg.

If your initial worksheet, the weight in kg stated is 1.
The unit for weight is not suitable for calculating a torque in Nm.
The unit suitable for this conversion is the gram. So it would be better to indicate on your worksheet that the weights are not 1kg but 1000g because your formulas use the gram as a unit of weight.
On your current worksheet, you are not using 1kg for the weights but 1g, hence the torque at 0, xxxx ...

But maybe you are already converting to g on your spreadsheet.

[harry ravi]

Have as much for me, the force of the weight this calculation with the unit of weight in kg and not in g.

Damage it would have been very useful.

This means that a larger wheel will have to be created.

40 weight of 200kg with squares of 1m side or a wheel of 14m high making a complete revolution in 1min:

-76,91Nm of torque
100kWh of power

It works less well but it works according to the calculations

[nlc]

Have as much for me, the force of the weight this calculation with the unit of weight in kg and not in g.

Damage it would have been very useful.

This means that a larger wheel will have to be created.

40 weight of 200kg with squares of 1m side or a wheel of 14m high making a complete revolution in 1min:

-76,91Nm of torque
100kWh of power

It works less well but it works according to the calculations

But no it doesn't work either!
If the small wheel is perfectly balanced why it would not be more by being larger and with heavier weights

[harry ravi]

I did not say that the small wheel is balanced, just that the torque is low enough, but there is a torque.

[nlc]

As explained sadist arnaud, you have a slight couple in a cyclic way to reach the position of equilibrium. It means as he explained very well that if you place it at a certain angle, it will move a little and oscillate around its closest equilibrium position, then stop.

He also explains very well that by modeling the wheel with squares and not trapezes, it is even worse the wheel is perfectly balanced whatever the position!

[Colmant]

Harry, your proto is he going ??

thank you Arnaud

[harry ravi]

No, my prototype is not progressing at all, I'm too busy nearby, I have to give myself a day to do it correctly.