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Effectiveness of mechanical power' multiplier power trains


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Testing so called windmill speed multiplier.

Below is the picture of *5.5*5.5= *30.25 multiplier power train:


I.e. vertical axle from the rotors is connected to the first large wheel - on the left - which gives *1 (normal speed). I.e. any consumer connected with an angled gear to this left wheel has *1 speed (NO multiplication).

Then second large wheel in the middle is connected to the first by an angled gear, and offers *5.5 multiplier. I.e. any consumer connected to the middle wheel gets *5.5 speed multiplication.

Then the third large wheel in the right/bottom is connected to the second by an angled gear, and offers *30.25 multiplier. I.e. any consumer connected to the bottom/right wheel gets *30.25 speed  multiplication.


Of course, all this train cannot move if not provided by enough power coming from multiple rotors. For sake of simplicity I assume all rotors are fully populated = each rotor has 5 sails sets applied.

Short explanation:

Once the multiplication power train is introduced, the final speed depends on the cumulative power of all rotors working in the train.

I.e. if you have speed multiplier of *5.5 - the torque/power is reduced to /5.5. If there's no enough power - the multiplier doesn't give you any advantage as due to low power the consumer will work with speed of *5.5/5.5=1.

There's a need to add torque/power by adding multiple rotors into the train.

And the number of required rotors is the central point of these tests.


So here are my empiric conclusions:

  1. Helve hammer has the highest inherent resistance. It is equal to ~1.2-1.3 rotors. I.e. in order to make helve hammer to work at the speed of the wind - you need more than 1 rotor. Actually 2 rotors.
  2. Quern has normal resistance equal to 1 rotor
  3. Pulverizer has low resistance equal to 0.7-0.8 rotors. It doesn't mean you will have it working faster, but it means that when working concurrently with helve hammer (1.2) it needs 2 rotors (1.2 of hh + 0.8 of pulv. = 2)
  4. Helve hammer has additional - kinetic - resistance, depending on the rotation speed of the toggle. The higher is the speed, the more resistance to the power source(rotors) it has.

Next conclusion is that it is better to have only one consumer working at the same time - several consumers slow the speed a lot. Use of transmissions and clutches is highly recommended.

It is correct for normal speed' power train, and even more correct and visible for multiplied one.


Assuming working with helve hammer as it is extreme of power consumption. Others are better.


For multiplier of *5.5 the user needs at least 4 rotors to have any effect, with wind starting from 50%. Otherwise it will be slow. Best amount of rotors for *5.5 it is better to have 8 rotors - full wind speed, multiplied by *5.5, will be applied on the helve hammer. 8 works always (almost).

For multiplier of *30 the system requires at least 28 rotors (7 large wheels with 4 rotors each), and in this case empirically the actual multiplication will be ~*15 at the most.

With *30 - the kinetic resistance is applied - which is clearly seen via calculation glitch: the speed falls to ~*6, then jumps to ~*20, then falls again to ~*6. All this due to the need to re-calculate the kinetic resistance once in a while instead of constant re-calculation. Bug?

28 rotors (7 layers of 4 rotors each) look like the picture below:



Major conclusion:

It is worth to have *5.5 multiplier as it solves the problem of blooms/blisters/plates as well as lime/flour/crushing.

Regular survival - 8 rotors on top of *5.5 multiplier - is enough for most cases.

For survival games, especially in single player - there's no need to have multipliers of *30 or higher. It is not worth it. The addition of *3 (from *5 to *15) requires 20 full rotors = 400 linen and tons of resin/fat.




I am intentionally skipping the point of having the lowest rotor layer at 60 blocks above the sea - this is already mentioned in wiki. But some funny point - wiki states that the height bonus is added to the "normal" speed, when it is actually multiplied. When having 0 knots - multiplication of 50% gives exactly 0 speed of the rotor. All because 0 knots multiplied by 1.5 (equivalent to 50% bonus) gives 0 (zero).


Thanks to Bogdan G. for inspiration.

Edited by BenLi
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