Re: Bitcoin puzzle transaction ~32 BTC prize to who solves it
⭐ Merited by citb0in (1)
you can run it as a lottery on rasppery pi or 1-core celeron cause it dosen't make any difference if you have 32 core cpu or 1 core cpu (in the second case you will for sure not find anything but at least you will not waste any money and electricity).
Why burn watts when you can harness Mother Nature? I’ve got a windmill powering a Raspberry Pi strapped to a seagull. Every time it flaps, that’s a hash. And if the seagull dives into the sea, I use a fishing rod, toss it into the ocean, and let the wind and fish solve the puzzle. Every time a fish bites, it's a new private key attempt. If nothing bites, at least you might catch dinner. Sustainable computing, baby! Solving Bitcoin puzzles with renewable energy and sushi. That’s a full brute-force sweep! Fish swim in binary, right? We just need to decode their wiggles. Solving Bitcoin puzzles, one gust and guppy at a time.
That's an awful lot of wasted energy. Just one hash? An RTX 4090 can do smth like 16 million hashes per a single joule.
Since I am, like everyone else around here, an expert in arctic wild life, let's clear things up.
Estimating the Energy Needed for a Penguin’s Single Wing Flap
Introduction:
To roughly estimate the energy required for a penguin to flap one wing once, we can use a simplified physical model. In this model, we treat one wing as a rigid rod rotating about an axis at the shoulder. The energy needed to accelerate this wing is given by the rotational kinetic energy formula:
Code:
E_rot = (1/2) * I * ω²
where:
I is the moment of inertia of the wing,
ω is the final angular velocity.
1. Estimating the Parameters
For a medium-sized penguin, we can assume:
- Mass of the wing: Approximately 0.2 kilograms.
- Length of the wing: About 0.3 meters.
- Moment of Inertia: Approximating the wing as a rod rotating about one end, the moment of inertia is estimated as:
Code:
I ≈ (1/3) * m * L² ≈ (1/3) * 0.2 * (0.3)² ≈ 0.006 kg·m²
- Angular Displacement and Time: Suppose the wing moves through an angle of 90° (approximately 1.57 radians) in about 0.2 seconds. The average angular velocity is then:Code:ω ≈ 1.57 / 0.2 ≈ 7.85 rad/s
2. Calculating the Rotational Energy
Plugging the estimated values into the formula:Code:E_rot ≈ (1/2) * 0.006 * (7.85)²
Carrying out the calculation:- *² ≈ 61.6
- Thus, E_rot ≈ 0.5 * 0.006 * 61.6 ≈ 0.1848 Joules (approximately 0.2 Joules)
3. Accounting for Muscle Efficiency
Since muscle efficiency in converting metabolic energy to mechanical work is typically around 20–25%, the actual metabolic energy required could be significantly higher. Assuming an efficiency of about 25%:Code:Metabolic Energy ≈ 0.2 J / 0.25 ≈ 0.8 J
4. Conclusion
In this simplified model, a medium-sized penguin might require roughly between 0.2 Joules (the calculated mechanical energy) and 1 Joule (when accounting for inefficiencies) for one wing flap. Note that these calculations are based on several approximations. The actual energy could vary with the wing’s mass, flap speed, angular range, and muscle efficiency. Also, while penguins use their wings in a “flying” motion underwater, the dynamics in water differ from those in air.