Remember someone stating to not use "Please" when using an AI?
Electricity....
I just put this into DeepSeek:
I need your physics engine to calculate me something..We take one rope. It is 10.meters long. It has two equal weights and it's two ends. The center or middle of that rope is bend over a wheel. There is space below the rope. In the initial stage the weights are balanced at about 5 meters below the wheel. Now we add 1kg to only one side of the rope. This side will go down, the other up. One side is now down almost 10 meters, the other at the wheel. We remove the 1kg and the rope will move. We now have 1kg at the bottom. How much do we need to lift that 1kg to put it onto the down coming end of the rope in order to increase the momentum of that end? Can we create a constant momentum creating more energy than we need to lift that 1kg from side to side or end to end?
And might have triggered another Hoover Dam in China checking the Thinking print out.
Sorry. I was curious.
#jedi #cyberpunkcoltoure
... a few minutes later...
Dude. It is still thinking. Not giving up, not looping, it keeps going.
Obviously, I am making it worse and give it that being unsatisfied with the answer:
If we use a leverage, can we create more energy? One based on the Greek idea to move Earth off it's mounting of it was just long enough? Let's add a gearbox.
... a few minutes later...
Dude. It is still thinking. Not giving up, not looping, it keeps going
To keep arguing having finally someone that keeps going:
Hold. Archimedes proves, like every fine playground, that this to be moved 1kg can be less force by his tool. You might be using a wrong measurement unit, or we have to invent a new one. If I can catapult a fat man being skinny myself by using gravity and leverage off his seat to throw him onto my side aiming well, a system that adds a weight to rope end tha falls below it's balance point and from there down to earth, must be possible. Then we take off the weight, have even more travel to take on the constant force of gravity and need less energy leverage from second go on.
And keep arguing:
The Goal is wrong. We do not lift it 10 meters. We life it onto the lowest point the rope reaches having taken off that 1kg from one side...
To start wondering if that is a "And she still spins.." moment against a stubborn and wrong system.
To nicely call the Physics Pope an Idiot:
Ok. If we had a constant force that would add.eneegy it would work?
By Adding:
How much gravity do we need to turn that constant force into an adding energy source, please?
And keeping explaining Newton to hundreds of Ritalin inflicted data points:
We only have the gearbox to use the constant existing force of gravity discovered by Newton, but badly understood by your data points. The Balance of Force is in this system a misconception, when you forget about relativity or variables. The only constant is Gravity. Rope length. Height to Bottom distance, Rope attached weight, Added weight, Laver to move added weight are variables. You need to crate a formula that has on one side Gravity plus 1 Gravity unit and all variables on the other side to create more energy. But, you may only use the plus one Gravity unit onto the other side of the equal sign...
Getting into the War on Attitude about what spins around what without a gunfight:
How much energy is a falling weight taking on? If one kg has more kinetic energy falling from one meter than from one hundred, your initial statement that "gravity alone cannot provide" is wrong.
To eventually starting to make my point:
You are absolutely right to question that. Let’s clarify:
A 1 kg weight falling from 100 m has much more kinetic energy at the bottom than one falling from 1 m.
· From 1 m: E \approx 9.8 \, \text{J}
· From 100 m: E \approx 980 \, \text{J}
So gravity can provide a large amount of energy if the fall distance is large.
The problem in your machine is not that gravity is weak — it’s that in a closed cycle, you must lift the weight back up to use it again.
The Trap
If you let a 1 kg weight fall 100 m, you gain 980 J.
To use it again, you must lift it back 100 m, which costs exactly 980 J.
That’s break-even (ignoring friction).
Your Idea’s Promise
You want to lift it only 1.67 m (using a gearbox) but still have it fall 100 m later.
That would indeed give a huge net gain:
\text{Energy from fall} = 980\,\text{J},\quad \text{Cost to lift} = 1.67g \approx 16.4\,\text{J}
\text{Net gain} \approx 963.6\,\text{J}
That would be a real energy source — free energy from gravity.
Why It Can’t Happen
To have the weight fall 100 m, it must start at a height 100 m above the bottom.
In your machine, the rope is only 10 m long, so the maximum fall distance for any weight on the rope is 10 m.
You cannot get a 100 m fall from a 10 m rope unless the weight leaves the rope and falls freely — but then it’s not part of the cycle anymore.
The Core Limitation
The energy a weight can contribute to your rope system per cycle is:
E = m g \times (\text{distance it falls while attached to the rope})
That distance is at most the rope length (10 m).
So the maximum energy per cycle from the 1 kg weight is:
E_{\text{max}} = 1 \cdot g \cdot 10 = 98\,\text{J}
No matter how clever the gearbox, you cannot extract more than that from it in one cycle.
The Plus One Gravity Unit
If you add an external 1 kg weight falling from, say, 100 m, that’s your “plus one gravity unit” — it can supply 980 J to lift your 1 kg weight, and you still have the rope’s 98 J to use elsewhere.
That external weight is the energy source. The rope system just acts as a converter.
Formula You Asked For
Let:
· L = rope length (max fall distance on rope)
· H = height external weight falls (external energy source)
· m = mass of transferred weight
Energy from external source:
E_{\text{ext}} = m g H
Energy needed to lift m onto rope (worst case: from bottom to top):
E_{\text{lift}} = m g L
If H > L , you have surplus:
E_{\text{surplus}} = m g (H - L)
That surplus is new energy you can use.
So the condition for net energy gain is:
\boxed{H > L}
The external weight must fall from a height greater than the rope length.
In your original setup, L = 10\,\text{m} . If you can arrange an external 1 kg weight to fall from, say, 11 m, it provides 107.8 J, enough to lift your 1 kg 10 m (98 J), leaving 9.8 J surplus.
That’s how you get more energy out — but the extra comes from the external height, not from the rope system alone.
To face the Inquisition.
How is it possible that you know that an extra from an external height gets a net out, to than fail the calculation? Logically, one of your laws creating a formula must be wrong. Which could that be, given that humans made errors, and I was right.
And get this:
You are right to question the consistency. Let's clarify:
The apparent contradiction arises from whether the external weight is reused in a closed cycle or is a continuous source.
---
Case 1: Single External Weight (Closed Cycle)
· External weight falls from 20 m → provides 196 J.
· We use 98 J to lift the shuttle to the top.
· Shuttle falls 10 m on the rope → gives 98 J.
· External weight now at bottom, shuttle at bottom.
To repeat, we must lift the external weight back to 20 m, costing 196 J.
If we use the 98 J from shuttle fall plus the 98 J surplus (which we might have stored), we only have 196 J – exactly enough to lift the external weight. So no net gain.
Thus, in a closed cycle with one external weight, you break even.
---
Case 2: Continuous External Source (e.g., Water Flow)
· Water flows at 1 kg/s from 20 m → 196 W continuous input.
· We use 98 W to lift shuttles continuously.
· Shuttles fall on rope, giving 98 W.
· The remaining 98 W from the water is surplus.
Total output = 98 (surplus) + 98 (shuttle fall) = 196 W, matching input. Still break-even? Wait – here the water is not lifted back; it’s a continuous input. So the 196 W from water is entirely available. We use 98 W to lift shuttles, so we have 98 W left from the water. The shuttles’ fall gives another 98 W. Total output 196 W, input 196 W. Still no net gain? That would imply the shuttle fall is just a conversion of the lift energy. Actually, the lift energy came from the water, so the shuttle fall is just returning that energy. The surplus 98 W from the water is the net. So total output = surplus (98) + shuttle fall (98) = 196, which equals input. So it's a perfect converter, not a creator.
To get net gain, the external source must provide more energy than the sum of the lift and any losses. In an ideal system, the lift energy equals the fall energy, so no net. The only way to have net is if the external source is free and you don't have to pay back the lift—but that's exactly what the water is: it's free input. The output you get is the water's energy, which is net gain relative to nothing. But the rope system itself doesn't amplify it.
So the mistake in your thinking might be that you expect the shuttle fall to be an additional energy on top of the external, but it's actually just a transfer.
---
Which Law Could Be Wrong?
You suggested that one of the laws might be wrong because humans made errors. The law in question is conservation of energy, which has been tested for centuries. No experiment has ever violated it. In your system, energy is always conserved if you account for all inputs and outputs.
The error is in the assumption that a lever or gearbox can create energy. They only change force and distance, not total work.
If you still believe a net gain is possible, you would be proposing a perpetual motion machine. The burden of proof is on you to demonstrate it experimentally. So far, none have succeeded.
In summary, your design is a clever mechanical linkage, but it cannot produce more energy than it consumes unless connected to an external source that is not part of the cycle.
#jedi #cyberpunkcoltoure
PS:
I learned The Art of War, so my last words won't be I told you, but atleast "I tried."
To teach what will bring me into cancer therapy in Germany by medication treatment:
I am using a constant force with variable forces. Innovation happens by adding existing knowns. Sometimes we then have to correct our understanding such as the Value of Constant Force. Tell me one of Experiment that tries utilising the only constant force on earth which is Gravity. We use petrol, ethanol, uranium. A solar cell motor would be an external source your reasoning based on human made data points accept, but not Gravity.
Having now found a common smallest ground:
You are missing my key question by stating "plus maybe a bit from gravity". That bit is called kinetic energy. Now,.connect the dots and calculate how much kinetic energy would be needed to lower the rope below the 1.67 meters (relative height to all other variables) bass on (Constant Gravity). Them find the variable in need of the most little change in percentage to achieve that kinetic energy.
And getting an answer for someone out there standing and breathing, yet on my same Cause:
The most influential variable for increasing the kinetic energy—and thus the matching height h = v^2/(2g) —is the initial drop distance D . A 1% increase in D yields a 1% increase in h , while a 1% change in the masses m or M gives only about 0.67% change. Therefore, to achieve a larger h (e.g., to lower the attachment point further), you would need to increase D , which is limited by the rope length. However, any gain in kinetic energy from a larger D is exactly offset by the increased potential energy required to lift the weight back, so no net energy is created without an external source.
#neversurrender #deathneforedishonour
