12 times more output than input, dual mechanical oscillation system !

[lancaIV]

Twice the speed halve the force.Halve the speed twice the force.
         Cube law ! Xv3 up or inverse cube down  !

+ 100% speed = + 700% force    -50% speed = -87,5% force

Base 1 +1 = 2 and 2v3= 8             1-0,5 = 0,5      (0,5x0,5x0,5)= 0,125

[Floor]

@ LankaIV

Interesting stuff !
              Lanka link...  cube law
https://www.quantum-controls.co.uk/insights/faqs/what-is-cube-law/#:~:text=Cube%20Law%20is%20the%20relationship,can%20give%20significant%20energy%20savings

"Cube Law is the relationship between a centrifugal pump or fans speed and its energy need is known as Cube Law. This means that a small increase in speed requires a lot more power, but also means that a modest speed reduction can give significant energy savings. A pump or fan running at half speed consumes only one eighth of the power compared to one running at full speed."

Correct me if I am mistaken.
I think this has to do with wind resistance or friction and its electrical energy
consumption,  not acceleration ?

https://phys.libretexts.org/Bookshelves/Classical_Mechanics/Classical_Mechanics_(Tatum)/07%3A_Projectiles/7.03%3A_Air_Resistance_Proportional_to_the_Square_of_the_Speed
... ... ... ... ... ... ... ... ...
Additionally it requires 4 times the energy to accelerate a given mass to twice its speed.

It requires 1/2 joule to accelerate a 1 kg object to a velocity of 1m/s.
It requires 2 joules to accelerate a 1 kg object to a velocity of 2m/s.
It requires 4 times as much energy to accelerate a 1 Kg object to twice the velocity.
Twice the velocity requires 4 times more energy (the velocity increase squared).
... ... ... ... ... ... ... ... ...
A 1kg object moving at 2m/s has 4 times the kinetic energy of a 1 kg mass moving
at 1m/s

[kolbacict]

If question 1 is how "much of a benefit is it at lower speeds".

I dont know. Does absolute speed matter?
p.s. If the original Milkovich double pendulum cannot be made self-powered, then I am thinking how to make it possible.

[lancaIV]

from conventional https://www.calculatorsoup.com/calculators/physics/kinetic.php


to
http://www.ijsrp.org/research_paper_feb2012/ijsrp-feb-2012-06.pdf


There are many complicated calculations and equations involved in understanding and constructing wind turbine generators however the layman need not worry about most of these and should instead ensure they remember the following vital information:


1) The power output of a wind generator is proportional to the area swept by the rotor - i.e. double the swept area and the power output will also double.


2) The power output of a wind generator is proportional to the cube of the wind speed.


Kinetic Energy = 0.5 x Mass x Velocity2 , where the mass is measured in kg, the velocity in m/s, and the energy is given in joules.


Air has a known density (around 1.23 kg/m3 at sea level), so the mass of air hitting our wind turbine. (which sweeps a known area) each second is given by the following equation: Mass/sec (kg/s) = Velocity (m/s) x Area (m2 ) x Density (kg/m3 ).
Therefore, the power (i.e. energy per second) in the wind hitting a wind turbine with a certain swept area is given by simply inserting the mass per second calculation into the standard kinetic energy equation given above resulting in the following vital equation:
Power = 0.5 x Swept Area x Air Density x Velocity3 , where Power is given in Watts (i.e. joules/second),
the Swept area in square metres, the Air density in kilograms per cubic metre, and the Velocity in metres per second.




or

https://www.researchgate.net/publication/330534847_Theoretical_calculation_of_the_power_of_wind_turbine_or_tidal_turbine


4.5 with "v2"  becomes total 4.6 "v3"  kinetik and potential energy sum effect




https://worldwide.espacenet.com/publicationDetails/description?CC=US&NR=2010283252A1&KC=A1&FT=D&ND=3&date=20101111&DB=EPODOC&locale=en_EP


[0090] Generator power and efficiency with wind turbine drive is computed below, for a representative example of the present invention, at maximum shaft speed, mid-speed, and minimum usable speed, using a few simplifying approximations. Shaft speed, power, and the other variables in the computations herebelow are exemplary, and not intended as limiting the present invention in any way. This will help explain FIG. 1 and FIG. 2 configuration operation, distinctions and improvements over the prior art.


[0091] Let maximum speed equal 1000 revolutions per minute (rpm), mid-speed equal 500 rpm, and minimum speed equal 100 rpm. Also, let maximum stator current Imax=10 amperes, and nominal VDC=100 volts. Further, let Q1-Q4 power MOSFET ON resistance Rdson=0.01 ohm, inductor L1-L2 winding resistance RL=0.1 ohm. Also, stator winding resistance R5=0.15 ohm, stator voltage Vmax=100 volts at 1000 rpm, and fly-back (free-wheeling) diode D1-D8 forward drop Vf=1-volt at 10 amp. These parameters are consistent with a test prototype, according to the present invention, developed to generate power from wind turbines.




[0092] At 1000 rpm, Vmax=100 volts, so PWM duty-cycle (Ton)/(Ton+Toff) is essentially zero. Therefore, losses Imax<2>(RL+Rs)+2 VfImax=(10 amp)<2>(0.25 ohm)+(2 volt)(10 amp), amounting to 45 watts loss. Output power=(Imax)·(Vmax)=(Imax)·(VDC)=(10 amp)(100 volts)=1000 watts.


Therefore, generator efficiency at maximum speed and maximum power is about 95% for this example of generator and integrated electronics parameters.


[0093] At 500 rpm, Imax=(10 amp)/(4)=2.5 amps; and Vmax=(100 volts)*(0.5)=50 volts. So PWM duty-cycle=1⁄2. Average pulse power generated=(Imax)·(Vmax)=(Imax)·(VDC)/2=(2.5 amp)(50 volt)=125 watts.


Losses to maintain inductor current=Imax<2>(RL+Rs+Ron)=(2.5 amp)<2>(0.26 ohm)=1.6 watts. Fly-back diode losses=2 Vf·Imax/2=(0.6 v)(2.5 amp)=1.5 watts. So total losses=3.1 watts. Therefore, mid-speed generator efficiency is about 97%.


[0094] At 100 rpm. Imax=(10 amp)/(100)=0.1 amp; and Vmax=(100 volts)/(10)=10-volts. So PWM duty-cycle=9/10. Average pulse power generated=(Imax)·(Vmax)=(Imax)·(VDC)/10=(0.1 amp)(10 v)=1 watt.


Losses to maintain stator and inductor current=Imax<2>(RL+Rs+2·Rdson)=(0.1 amp)<2>(0.27 ohm)=0.0027-watt. Fly-back diode losses=(2·Vf)·(Imax)/10=(0.6 v)(0.1a)/5=0.012 watts. So total losses=0.015-watt. Thus, generator efficiency at low speed is about 98%.




https://www.city-journal.org/wind-power-is-not-the-answer
But the new study, published in Environmental Research Letters, shows yet again that wind energy's Achilles heel is its paltry power density.


"We found that the average power density—meaning the rate of energy generation divided by the encompassing area of the wind plant—was up to 100 times lower than estimates by some leading energy experts," said lead author Lee Miller, a postdoctoral fellow who coauthored the report with Harvard physics professor David Keith.


The problem is that most estimates of wind energy's potential ignore "wind shadow," an effect that occurs when turbines are placed too closely together: the upwind turbines rob wind speed from others placed downwind. 






It is called

BROAD SPEED RANGE GENERATOR


it can be also coupled,instead with wind blade/s ,with a compressed air engine,an ic engine or electric motor !


But the speed/power ratio under "inverse/cube law condition" stays valid !


Attention : above 1000/500/100 RPM " PWM duty-cycle " numbers


For the motor ,from DC or AC to Pulse ( PM or capacitive) : https://de.scribd.com/document/394252850/The-New-Law-of-Electrical-Power-Formation


Is in the papers an inverse/square law or inverse/cube law argumentation to find ?
http://www.veljkomilkovic.com/Naucni_radoviEng.html

[Floor]

If question 1 is how "much of a benefit is it at lower speeds".

I dont know. Does absolute speed matter?
p.s. If the original Milkovich double pendulum cannot be made self-powered, then I am thinking how to make it possible.

Same here.  I'm Just looking at various approaches.  I don't see the use of this double pendulum
as being something that will improve the power in to power out ratios.

But Lanka sure did popup some interesting links !

   best wishes
        floor