# Dividing torque of one axle to three in varying proportions [closed]

I want to power a small mechanical monstrosity, which performs three tasks that require three separate axles. For the powering, there's a hand-operated crankshaft connected to the primary axle that has 3 gears located at (equally-distanced, if that's a variable) intervals and there will be 3 secondary axles, each with their ends connected to the 3 gears on the primary axle respectively.

I want the TORQUE of the primary axle to be distributed in a 10:30:60 ratio to the secondary axles.

Which law can I use to calculate the size and number of teeth for the gears on the primary axle, as well as their mating gears on the secondary axles each? Please consider I have little knowledge pertaining to gears, apart from what torque, mathematical ratios, forces, distance, and other basic physics quantities are.

EDIT: For those who think Swiss machines and god-knows-what apply to my question, HERE is my idea:

Maybe I am using the wrong terminology in the question.

All pumps are identical (screw) pumps pumping water, the flow rate must be highest in pump 3 and lowest in pump 1, while pump 2 has a median flow rate. The flow rates are in the 10:30:60 ratio for pump 1:2:3 respectively.

Now I think that to have a higher flow rate, we need more revolutions per minute for the pump, but that requires more torque (or doesn't it?). Basically, in the end I need pump 3 faster than pump 2, and pump 2 faster than pump 1. Whichever way you come to it, be it so.

If I need more gears to achieve this, then I'd be happy to have a diagram to show me where and how.

• Are you talking about a manual powered Swiss CNC machine? Also, you want do you want a 1:10, 1:30, and 1:60 power distribution to axles 1, 2 and 3- or 10:60 (1:6) to each axle?
– jko
Jun 10 at 11:50
• Possibly helpful: khkgears.net/new/gear_knowledge/… . But your question is unclear: do you need to transfer power in those ratios but you don't care about the speed of the output axles? Jun 10 at 12:11
• @jko Hell no Bro, what on earth is a Swiss CNC? Also, I want a 1:10, 1:30, and 1:60 power distribution to axles 1, 2 and 3- Jun 10 at 13:38
• @carl-witthoft I need to transfer powe in those ratios, NOT speed. Jun 10 at 13:40
• Are you sure you don't mean torque instead of power? It is much easier to separate power electrically than mechanically on a fixed axle system. A "Swiss machine" was a precursor to modern CNC style machines, would behoove you to check it out since it sounds similar to what you're asking about.
– jko
Jun 10 at 15:35

UPDATE

Centrifugal pumps generally obey what are known as the pump laws. These laws state that the flow rate or capacity is directly proportional to the pump speed; and the power required by the pump motor is directly proportional to the cube of the pump speed. These laws are summarized in the following equations.

$$\dot V \propto n \tag{eq.1}$$ $$P \propto n^3 \tag{eq.2}$$ where:

• $$n$$ = speed of pump impeller (rpm)
• $$\dot V$$ = volumetric flow rate of pump (gpm or ft3/hr)
• $$P$$ = pump power (kW)

Since the flow rates on pumps 1,2 and 3, have a ratio of 1:3:6, then the

• the rpms on each pump $$n_1,n_2,n_3$$ will need to be 1:3:6

• Power requirements ($$P_1:P_2:P_3$$) for each pump will be $$1:3^3:6^3$$

• the Torques on each pump shaft $$M_1, M_2, M_3$$ will need to be $$1:3^2:6^2$$ (because $$M=\frac{60 P}{2\pi n}$$)

What drives the Power and the Torque is the rotational speed of the pump shaft. So, each pump shaft need to rotate at

$$n_1:n_2:n_3= 1:3:6$$

Given the above relationship you need to use gear diameters $$d_1:d_2:d_3 = 6:3:1$$

I think you are mistakenly assuming that if you use different size gears, somehow different torque will be applied to each gear.

To simplify things, let's take a shaft with only two gears

Assuming the shaft has a power P at $$n$$ rpm, then the torque supplied will be:

$$M_t = \frac{60 P}{2\pi n}$$

The torque will be divided among the two gears 1 and 2.

• I will take the extreme, when gear 2 has nothing connected on it. In that case all the torque will be directed at gear 1.

$$M_1= M_t$$ and the tangential force on the gear 1 will be equal to

$$F_1 = \frac{M_t}{r_1}$$

Where $$r_1$$ is the radius of gear 1.

• Similarly, when there is no load on gear 1, then all the torque will be on torque 2. $$M_2= M_t$$

• In the general situation that $$M_1$$ is the torque on gear 1, then the available torque for gear 2 is $$M_2= M_t- M_2$$

• if for some reason $$M_1 + M_2 > M_t$$, then the shaft will start to decelerate (kind of like stalling).

Bottom line:

the torque distribution depends on the load of each gear.

In order to distribute the torque( or power) at a fixed proportion on a single shaft axis, you need special conditions. The way you have your setup - at least as much I understand it- you will be able to control the speed on the other shafts/gears but not the gear distribution.

I don’t think can be achieved by gears only.

The best solution depends on what your goal is, if the motor is capable of feeding al the pumps then the system will balance itself powering all the pumps with the torque they need depending on the load they are subjected to.

If you need this kind of device in order to be sure one pump isn’t drawing all the power from the energy then I would go for torque limiting joints, they usually rely on friction and springs to limit the torque that is transferred through them. You can’t achieve a fixed ratio distribution independent of the input torque so you should set the joints on the absolute value of torque not on percentages