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¿ Why trying to find the complex way to do things ? You want a spark plug to spark, use a spark plug transformer from a car, motorcycle, lawnmower, whatever. Apply 12 VAC to it and watch where you stick your fingers. Miguel

Do not use polyestyrene foam as packaging. Use sponge instead. Only 50 metres? -piece of cake!- Helium balloons will do it, but there is risk the 50 metres can become 50 kilometres and get the first prize plus too many extra bonuses. A tiny leak with an aquarium hose valve on the helium balloon can be set for shorter distances and a very smooth landing and may need very little protection. If a strict path has to be followed by the rules ot the contest, a second air ballon can provide propulsion to the helium carrier while riding captive on a string from start to goal line. Miguel

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-----------> http://www.scirus.com/srsapp/

Very interesting, Samphire. If the tiny currents the neurons carry make equally tiny magnetic fields, varying magnetic fields should produce currents in them ! Then, the magnetic fields of soooo many electronics surrounding us should induce stray currents in our neurones, interfering with the nervous system, making us... crazy ? Miguel

JC1 : You will really, really bang yourself your head against a wall to bleeding when you grow up and read your post a few years from now. Miguel

Many pictures on the news show the discharge ends of the pump pipes at 10 -if not more- feet above the surface. The brutal flow reduction and increased amount of fuel makes no sense. And that pumping operation is conducted by "experts" ? The discharge ends should be as close as possible to water level to be efficient, probably doubling flow. SHAME ! -The land of waste- Miguel

Hi. The first; what size of a mylar mirror "umbrella" held by a reasonably far positionable satellite could produce a ~500 Km Ø shadow on earth? And the other; would such shade cool the sea surface a few degrees if projected several days on a path, enough to somewhat decrease the intensity of a hurricane ? Miguel

Hi. I left out a factor in the formula, am sorry. The correct one is here: http://www.imagineeringezine.com/e-zine/capacitance-3.html Miguel

70 feet² = 65032 cm²

Hello. The formula is: C in Farads = Plate area in cm² x k (dielectric constant) ÷ distance between plates in cm. The constant for waxed paper is around 2. And good for up to about 300Volts. Miguel

Hello xeluc. ...."I understand everything you said; You said a 1 Farad capacitor charged with 9 volts can discharge 1 Ampere in 9 seconds or 9 Amperes in 1 second. How would you pick which one you wanted ot happen, a resistor?" --------> Yes, the resistor . But your wording should say {Charged TO 9 Volts, not WITH} A 1 Farad capacitor charged to 1 Volt would have: 1 F x 1 V = 1 Coulomb. If the wires discharging it have 1 Ohm resistance: 1 V ÷ 1 Ohm = 1 Ampere and the time to discharge is 1 Coulomb ÷ 1 Ampere = 1 second The other case, with the voltage being 9 Volts: 9 V ÷ 1 Ohm = 9 Amperes and the time to discharge is 9 Coulombs ÷ 9 Amperes = 1 second Or; if you increase the resistance of the discharging wires With the resistance being 9 Ohm instead of 1 Ohm, 9 Volts ÷ 9 Ohms = 1 Ampere The discharge time is : 9 Coulombs ÷ 1 Ampere = 9 second [if you increase the resistance, the current will discharge in a longer time] Remember: Coulombs ÷ seconds = Amperes = Volts ÷ Ohms And imagine a swimming pool (large CAPACITY capacitor) filled with 2 inches of water (~2 Volts) and a frying pan (small capacity capacitor) filled also with 2 inches of water (same ~2 Volts) The time needed to empty each trough equal drain pipe size (resistor) will be different. Changing the size of the drain pipe (resistor) will change the draining times. The water depth is the voltage; the size of the container is the capacitor capacity, the restriction by the drain is the resistance. I want to correct MY mistake in post #20 above. Where says " Energy in Joules = voltage x capacity " Should say "Energy in Joules = voltage x charge" Miguel

Hi xeluc. (This is long because you made it that way) ..... "my questions are about the current produced when the two wires of a charged capacitor are brought together. Since noone is touching my huge paragraph, ill structure two questions I have in Paragraphs. -----------> The current produced joining the wires of a charged capacitor is : Amperes = charge in Coulombs ÷ time in seconds. And the Amperes will be the Volts ÷ by the resistance of the wires. Then, Coulombs ÷ seconds = Volts ÷ Ohms Question 1: If I have two different sized capacitors and they are both charged using 1.5 Volts DC; What difference would there be in the Capacitors. The larger one would obviously Have more electricity in it, but would the larger capacitor have more volts or Amperes in the current it produced when connected to itself. ----------->The larger CAPACITY one (not size !) will have more charge in Coulombs, (not volts nor amperes) and (not electricity) Example: If the large capacitor is 1000 µF, charged to 1.5 V, it will contain 1500 µCoulombs. If the small capacitor is 10µF, charged to the same 1.5V, it will contain 15 µCoulombs Question 2: Do capacitors dump out all of their energy at once or is it relativly slow (by that i mean a few seconds/minutes instead of instantaniously)? ---------> the time to discharge to 66% is time in seconds = Charge in Coulombs ÷ current in Amperes. It doesn't make much sense for a computer to use a capacitor if all energy it had is expelled instantaniously, it woul have to be able to slowly draw power from the capacitor. So either the capacitor dumps it's energy slowly or there are other components that conserve the charge. Can soemone please tell me which? Because Camera flashes use capacitors that dump their energy all at once.. ---------> Yes, but you are still confused. A camera flash will discharge the capacitor in few milliseconds. And will take many seconds to recharge. And a capacitor in a computer circuit is NOT meant to discharge and recharge cyclically; it is meant to discharge briefly a small amount of its charge while continuosly being charged ALSO. So, if I had an extremly large capacitor, say The size of a computer tower, and hooked up a variable resistor to it, I could make a low voltage that would last a while or a high voltage that would go by quick.... right? If so, this serves my purpose exactly. ----------> It would not MAKE a lower voltage, it would discharge a CURRENT. A larger CAPACITY capacitor (NOT SIZE!) will discharge slower trough a larger resistor. Lastly, The two capacitors are charges with 1.5 volts. I'm just assuming that you "short" the capacitor. So we'll jsut say that the resistance is negligible. SO that's all there is in the circuit. My question would be, if i hooked up a voltmeter to it would it read 1.5 volts or would it read a higher voltage the bigger the capacitor is. -----------> both capacitors charged to 1.5 Volts, shorting the leads, the voltage will decrease immediately in microseconds. Same thing with Amperage. I thought that capacitors kept a constant voltage ------------> NO. Voltage nor Amperage is NOT constant during discharge nas the amperage dereased and an inductor keeps the same amperage while the voltage increases and decreases depending on the power going to it. Am I right or wrong in saying this... ------------> You are confused. So what your saying is that If i charge the capacitor at 1 volts. It will discharge at 1 volts with a varying amperage? -------------> No. The voltage will start decreasing immediately when current starts flowing. So it could be said that a larger Capacitor would hold a higher "start" amperage, therfore holding the voltage longer... right? -------------> Yes. Does that also mean that you can store more energy in a capacitor at a higher voltage? -----------> Yes. Energy in Joules = voltage x capacity. More Volts = more energy If you can charge a capacitor 1 Ampere at 1 volts and 1 Ampere at 20 volts, then the only real limit to the amount of charge the capacitor can achieve is either when theres so many electrons in the capacitor that it physically cant fit more in it or if the voltage gets so high that it spontaniously arcs and shorts itself out... I think this is all right.. yes? no? If all this is true then my questions are answered! ----------> No and yes and no. You do not charge Amperes to a capacitor; you charge Coulombs. The limit is when its capacity is filled. It will arc if its voltage ratings is exceeded. your EDIT: I guess when I say 1 ampere I really mean that the capacitor is rated at one farad, so the capacitor would discharge 1 amp in 1 second at one volt. -------------> If you meant 1 Farad charged INITIALLY to 1 Volt,( = 1 Coulomb) it will discharge 1 Ampere during 1 second. But if the capacitor were charged at 9 volts, the nthe capacitor should discharge 1 amp at 9 volts in a second. OF COURSE this does not take resistance into effect; All I'm trying ot do here is understand capacitors fully, I'll add resistance to the equation later ---------> If 1 Farad is charged to 9 Volts, (9Coulombs) it can discharge 1 Ampere during 9 seconds or 9 Amperes during 1 second Ok, that's great. So is there any way of... inferring what the voltage and current would be? There has to be some ratio or equation.. ----------->The current is = Coulombs ÷ time in seconds BTW, thanks for taking the time to help me out. It IS appreciated your edit: I think im understanding this. The size of the capacitor has nothing to do with what the voltage and current will be, the size will determin how long that charge will last? -----------> You are getting closer. Suggestion : Print this and read and re-read slowly. Miguel

Hi. Use a VCR. Connect your camera signals to video and audio inputs, and feed your TV ANT IN from the RF OUT threaded connector. If you know some electronics, you could canibalize the modulator inside a discarded VCR. It is the tin box inside it that has the RF OUT connector attached to it and a channel 3/4 selector on it. It will; need about 8 to 12V to energize it. Miguel

If I understand your question, to concentrate HEAT, you must have a larger MASS holding the heat If you want to raise/concentrate TEMPERATURE, a focusing array of mirrors could do it. If the heating element in the chamber is an electric resistor, a parabolic/eliptic mirror behind it will concentrate the temperature to a focal point. Miguel

As expressed above, you are lost in space about what a capacitor is and how it works. After charged, as there is no current, there is no magnetic field. There is only electric field. If you want to do your experiment, buy a vacuum capacitor and stick a magnet outside. Then you will have a magnetic field between the plates. Miguel

....."First off, I need clarification on some things. I know what an amp is. It's how fast the current is moving. Great. ------> WRONG. First, it is not amp. It is Ampere, and second, is the amount of electrons flowing per second. EXACTLY as water flowing in a pipe at a constant velocity. Now Voltage. I know the literal definition. It states that Voltage is the Potenital difference of charges or something ot that effect. --------> your "something ot that effect" Is like the PRESSURE on a water pipe. What I may or may not understand is it's relationship with Amperage. --------> More Volts, -> more Amperes. As if the water has more pressure, more amount of water will flow. The rest of your post is a salad of confusion, set it aside until you understand the concepts clearly, and then we can continue ONE question at a time I know that you can convert high voltage low amperage currents to low voltage high amperage currents, but I don't understand how voltage is measured. It's easy to understand that current moves faster in higher amperage currents so that means high amperage can kill you easily. thing is, If I have a high and low voltage of the same amperage going throuh me, what would be the difference. I realize that the higher voltage could be converted to a higher amperage, then again so coudl the lwoer voltage. So like I said, whats the relationship here, other than jsut sayuing voltage is lie kthe pressure in a water pipe. that doesnt hlep. What difference does a high and low voltage have when both currents have the same amperage? So that's that. Now we go to resistance. I understand the equation I=V/R . It's all good. Here's what I dont get. If you have a current thats 9 volts flowing at 1 amp, the resistance would be 9 Ohms, correct? That would satisfy the above equation, but then you have to deal with the resistance of the wire the current is traveling through. I understand that the resistance would be higher (since the equation could be re-arranged as IR=V) so wouldn't the voltage continually rise as it flowed through the wire? I know it doesn't SEEM correct, but that's what the equation states. Of course some power would be lost due to heat from the resistance, but I'm not sure if that's relevant. So here comes my last 2 questions. Sorry again. If I have 2 capacitors; one the size of a marble and the other the size of a 2-liter bottle, and they were both charges at 1.5 Volts; What would happen? Would they both be charged at 1.5 votls and the bigger capacitor have a larger amperage? --------> The charge on a capacitor is not Volts nor Amperes, it is Coulombs, equal to Volts times Farads. Would they both have the same amperage and the bigger capacitor have a larger voltage? Would the larger capacitor have a larger amperage AND voltage? If so, what determines the ratio of power between voltage and amperage. So thats one quesiton ; here is the last one. If I have 9 volts flowing at 1 amp from a capacitor, and in the "circuit" is a mor eresistive material, like graphite; would that raise the voltage since resistance was added? That ties in with a previous question I guess.. So thats everything I'd lie kto know. I'm sorry if all this has been answered, i looked around for a little while.. And I'm sorry that this was very long also. SAo thanks in advance for anyone who helps me!

Hi. Just develop an experiment to measure the propagation speed of gravity.

To make things harder, I don't think so ! A water balloon, and a plain air balloon, both identical and inflated to the exact same size, will be affected by the same air drag resistance, and will have different weights. For very sure, launch both from your Empire State building, and you will find that a long time after that pedestrian dries out, the air inflated one is still flying. Even if you do it in your bedroom with zero crosswind, they will not fall at the same speed. It's the law of selective gravity. Miguel

Thanks, gentlemen. From the responses above, the list of reasonably pure abundant natural soils could be: Lime Chalk Salt Coal Silica Water Silicates of calcium Silicates of aluminum Silicates of magnesium Silicates of iron Sulfates of calcium Sulfates of barium Sulfates of strontium Chlorides of calcium Chlorides of sodium Chlorides of potassium Chlorides of ammonium Calcium fluoride Nitrate of ammonium Nitrate of potassium Now, here is where you, skilled guys, come to play: Using any of the above, or more than one, plus heat plus any other requiered compound if needed, what mineral fuel can you come up with ? Example: Coal + lime + heat = calcium carbide -yields acetylene fuel- Let's get rid of petroleum, yes ? Miguel

Which soils are abundant in somewhat pure form in nature ? Like lime, chalk, sand, what others ?

Hi. Which drag coefficient is greater; a rounded leading edge or a razor sharp leading edge for a given wing/keel ?

NO. Homework was 30 years ago.

Hi. What is the force exerted on a 1m² vertical plate held transversal submerged in a river flowing laminar at 0.5m/s ?

Perhaps was improperly expressed... Would the words WATER WHEEL make more sense than paddle wheel? Miguel