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elementcollector1's Achievements


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  1. I ended up trying this again, this time using radiation physics for a different approach. This time, I assumed Magneton and its surroundings had reached thermal equilibrium, so time is no longer a factor (but is presumed to be quite a while). I started by getting the solid angle of an area 1km from Magneton, assuming Magneton could be expressed as a 1.0m diameter sphere. This got me an angle of 3.1415 x 10-6 steradians. From here, I (maybe correctly?) asserted that because Magneton and its surroundings both started at the same ambient temperature, the temperature added by radiation could be estimated via the following equation (taken from here) and used to solve for Magneton's surface temperature (again added to the ambient temperature): ΔT = +2 K = ((ΔTMagneton)4 (3.1415 x 10-6) / (4pi))1/4 => ΔTMagneton = 89.45 K I then went back and altered the solid angle's distance to get the temperatures at 750, 500, 250 and 1 m from Magneton using this same ΔTMagneton, and graphed these to find the following equation: ΔT = 63.251d-1/2 Where d is the distance from Magneton in meters. Integrating this from d=1 to d=1000 to find the area and then dividing it by the distance of 999 m between the two points, I found the 'global' temperature increase within the 1 km sphere to be an average of 4.004 K, which seems pretty reasonable. Lastly, we can add this temperature increase to the average global temperature (I picked 1998's value of 58 oF or 287.594 K because this was when the game came out) and use the Stefan-Boltzmann Law to calculate the energy density in Joules/cubic meter of the 1 km sphere Magneton is inside, followed by the total energy in Joules to maintain this temperature inside the 1 km sphere. pE = (287.594 + 4.004)4 (5.6625 x 10-8) (4/(3 x 108)) = 5.459 x 10-6 J/m3 E = (5.459 x 10-6 J/m3) (4/3)(pi)(10003) = 22,866 J This is a surprisingly small amount, which I think makes sense at thermal equilibrium, but why is it so much different than the one calculated previously? Also, would I be right in saying 23 kJ/s or 23 kW is needed for Magneton to maintain this temperature increase every second? So not only is it hot enough to boil water on contact, Magneton is also pumping 23 kW of microwave radiation into its surroundings. Anything nearby is going to get cooked pretty quickly (temperature at 1 m under these assumptions was +63.25 K, meaning a nearby Trainer or Pokemon is in a lot of trouble unless they have RF shielding...)
  2. Spotted this playing one of my old Pokemon games under the Pokedex entry for Magneton: "It is actually three Magnemite linked by magnetism. It generates powerful radio waves that raise temperatures by 3.6 degrees F within a 3,300-foot radius." That got me wondering how much power that would actually need, and whether it would be feasible for Magneton to actually output it via microwave radiation. Best equation I could find was the heat capacity equation (altered for power instead of energy): P = mcΔT/t (P = power in watts, m = mass in kg, c = heat capacity in J/kg.K, ΔT = temperature change in Kelvin, and t = time in seconds) We know ΔT is 2 K, and I'm assuming c is the value for water vapor (steam) at 1996 J/kg.K. m is a bit trickier - I assumed an average relative humidity of 30%, which gives 0.0066 kg water vapor per cubic meter air. In non-Imperial games, that 3300 feet is replaced with 1000 meters, so I'll use a sphere of 1000 meter radius for Magneton's maximum influence, which has a volume of 4.1887902 x 109 m3, giving a total water content of 27646015.3 kg. t is also tricky - I assumed Magneton takes one hour (or 3600 seconds) to achieve this temperature rise. No real reason behind the assumption other than giving it plenty of time. From all this, I get a value of 3.0656359.2 x 107 W, which seems pretty realistic given the sheer mass of water being heated (and pretty terrifying for a lone Pokemon to be outputting somewhere in the wilderness). However, I want to take into account more complex factors, such as attenuation of the radiation, the frequency of the microwave (which I'm assuming is 2.45 GHz, since that's the one commonly used in commercial microwave ovens). I found this graph for attenuation of a range of EM frequencies from oxygen and water: However, I'm not quite sure how to factor all this stuff in. Plus, assuming the number doesn't drop too much, what would be the temperature one meter from Magneton? What would be some other consequences of that much microwave radiation in the atmosphere?
  3. I recently read about an interesting and potentially relevant housing development: Installing packets of paraffin wax in the walls. These absorb the heat during the day, and release it at night, lowering the change of temperature in rooms. Found here: http://www.wpi.edu/Images/CMS/News/Apelian_JOM.pdf
  4. Sn's valency is 4, not 6 - the Na+ ions take care of the rest. As for separation, I would suggest mixing with an acid to change the pH - hydrous Sn(OH)2 or Sn(OH)4 (can't remember which) should precipitate in a certain pH range.
  5. Not sure what you mean, Enthalpy - I've been doing calculations this whole time. Anyway, if there's no way to "cheat" the work equation (W = Fd) into thinking there's more distance than there really is, I think I'll have to settle for a system with a lower tolerance for falls - better than what a human could manage, but not perfect.
  6. Hmm. But shouldn't the plane be heavy enough that the reduced vertical momentum (from the action of the wings) is still greater than the vertical momentum of a person at terminal velocity? Source: http://www.avweb.com/news/savvyaviator/192153-1.html?redirected=1 3g might not be much acceleration, but considering the plane that's a whole lot of force. Then again, if deceleration's the issue, wouldn't there still be a problem even if all the force of landing was absorbed? Back to your earlier calculations on deceleration, the minimum distance to get the deceleration within "safe" (relatively speaking) limits would be 12 m. Is there any way to make this length less - as in, could it be coiled up or otherwise rendered shorter while still acting as if it were the full length?
  7. I don't get it - if the strut can stop an airplane, why wouldn't it stop a human (ignoring practicality)?
  8. Well, unless time travel is involved, I don't think I'll be applying the second method. After some thought, I wondered why we could survive airplanes landing (after all, that's quite some momentum!), and looked it up - apparently something called an 'oleo strut' absorbs most of the shock, apparently by damping with compressed air or nitrogen and oil. Hypothetically, what if a human pogo-sticked one of these off of a 1200-meter cliff (or some similar setup where the strut would take the brunt of the impact), in keeping with the earlier numbers - would they still be injured from deceleration? I would think they wouldn't if a few of these things could stop the vertical momentum of a passenger plane, but maybe I'm missing something more subtle.
  9. So if the acceleration is directly proportional to the stopping distance, what is the upper limit on deceleration for a human? Also, would it be possible to 'transfer' the acceleration somewhere else - as in, direct the shock somewhere it can be less harmful? Not sure on how shock 'moves', at any rate.
  10. So why is an airbag able to decelerate a person safely? After all, it's not all that big compared to a person, yet it stops collisions very well and allows the driver to survive mostly intact.
  11. What would be an appropriate "soft" substance? Foam, or maybe a type of rubber?
  12. Huh. So energy and shock absorption do nothing in this case, because the deceleration itself is enough to damage the human body?
  13. Okay, so assuming two pads, one on each of the person's feet, the contact area for the body would be about 20 cm by 10 cm, or 200 cm2. How would the force exerted on this area be calculated, then?
  14. It's mentioned on the webpage (http://www.sorbothane.com/) that it can be used for both "shock" and vibration applications.
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