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About DandelionTheory

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  1. in the attached picture i attempt to show the force I'm working with. The idea is to exploit the force on AB. although the currents in the example DO NOT CREATE A CIRCUIT, they show the force im working with. IF, the circuit were to be closed with a wire from D to C, the total force on DC would cancel out the total force on AB. also, We find the magnetic field vector direction by using the right hand rule. Because the current vector is inline with the writing surface, the magnetic field vector is represented as an X for into the surface and O for the vectors outward direction. Magnetic fields at 90 angles from an ions velocity vector do something curly to its path, but still follow electric field lines. not talking about radiation. im talking about magnetic fields acting on currents in wires and free ions. change your drugs bro.
  2. You don't understand what I'm doing, first understand what I'm saying then you can maybe do something about correcting it. What happens to each current-carrying wire? the total force on each current-carrying wire is in the positive y direction. Some vector directions cancel out.
  3. Man that would suck if they weren't in a circuit. Yet..... Maybe, maybe if it was in a vacuum. Look at current direction.
  4. I did, with little blue F's So how are ideas shared? Is it with words and numbers? Can I use reference? It's condescending to tell me to reference something when I referenced it. I would like some support communicating with you, but I'm having trouble. Do I really need to kiss you? Do I have to ask directly or do you just delete people that have trouble communicating? Bro get up, get your big boy pants on and teach me.
  5. Not for accelerating free ions. FORCE ON CURRENT CARRYING WIRE
  6. FORCE ON A CURRENT CARRYING WIRE. ITS CALLED THE LORENZ FORCE. I honestly need to put it down? Can you not read or reference it yourself? I don't need to back up a WELL KNOWN CALCULATION STRANGE. grow up.
  7. How about without permanent magnets?
  8. What say you smarty pants? I was way into the impossible em drive, smoked a huuuuge bowl and discovered the basic idea. If the hypotenuse of a right triangle is open for charge carriers to be free, force applied to the system from the hypotenuse can be negated due to it not being physically attached. If 2 "circuits" are combined at the adjacent leg, the force on each of those legs is cancelled out in the system. What's left is force on the opposite legs. -DT Lol #winning https://www.reddit.com/u/DandelionTheory?utm_medium=android_app&utm_source=share
  9. I've been trying to put the math together for this, its out of my league. What i need to answer is: if each charged plate strips/adds electrons to the air molecules they touch, will this configuration of insulators and charged plates produce positive thrust along the main Y axis? -DT PS, "Tah-Dah!"
  10. i have attached a picture attempting to show you some detail. in the picture it depicts wire configurations to exploit the lorentz force on a current carrying conductor so the force vectors on part of each current loop are up. ive been attempting to use current loops to get this to work, but its not much more than a rail gun... the question i have is: aside from the rails repelling each other in a rail gun perpendicular to the armature vector, is there any reason the armature cannot be welded to the rails and enough power applied to force the whole contraption off the ground; and not shoot off like a bullet?
  11. i made a mistake Lxx = length of the section of coil C1 being acted on. also the integrals need to be edited, i didn't have the presence of mind at the time to see this mistake. i'm glad i have a friend to point it out. i have corrected the following variables because the second post cannot be edited any longer.: FnetAD = dFAJ = I1dLAD x (μ0I2/2πdr) FnetBC = dFAJ = I1dLBC x (μ0I2/2πdr) FnetAB = dFJK = I1dLAB x (μ0I2/2πdr) FnetCD = dFJK = I1dLCD x (μ0I2/2πdr) FnetAD = dFDK = I1dLAD x (μ0I2/2πdr) FnetBC = dFDK = I1dLBC x (μ0I2/2πdr) FnetAB = dFAD = I1dLAB x (μ0I2/2πdr) FnetDC = dFAD = I1dLDC x (μ0I2/2πdr)
  12. I have a solution. With respect to the picture i attached to this reply, you have to calculate the net force of coil C2 on coil C1. if we assign points, with respect to the rules in the OP, we can assume coil section JK of C1 will experience a greater force from coil C2 section AD than coil C2 section BC will experience from coil C1 section JK. buuuuut you want math, so... to calculate net force coil C2 induces on coil C1 you must brake up each side of coil C2 and calculate its force on each side of C1. so the total net force of coil C2 on coil C1: Ftotalnet = FnetAJ + FnetJK + FnetDK + FnetAD now for a perpendicular current we need integrate, because the farther you move from a magnetic field source the weaker it becomes. to calculate the force on each section of coil C1 when the current section of C2 is perpendicular, we have to break up the length into arbitrary lengths d, calculate the force on each length, and add the net force. FnetAJ = FnetAD + FnetBC + FnetDC FnetAD = dFAJ = I1dLAJ x (μ0I2/2πdr) FnetBC = dFAJ = I1dLAJ x (μ0I2/2πdr) FnetDC = I1b(μ0I2/2πrAJ) FnetJK = FnetAB + FnetCD + FBC + FAD FnetAB = dFJK = I1dLJK x (μ0I2/2πdr) FnetCD = dFJK = I1dLJK x (μ0I2/2πdr) FBC = I1b(μ0I2/2πrJK) FAD = I1b(μ0I2/2πrJK) FnetDK = FnetAD + FnetBC + FAB FnetAD = dFDK = I1dLDK x (μ0I2/2πdr) FnetBC = dFDK = I1dLDK x (μ0I2/2πdr) FAB = I1b(μ0I2/2πrDK) FnetAD = FnetAB + FnetDC + FBC + FAD FnetAB = dFAD = I1dLAD x (μ0I2/2πdr) FnetDC = dFAD = I1dLAD x (μ0I2/2πdr) FBC = I1b(μ0I2/2πrAD) FAD = I1b(μ0I2/2πrAD) where: I1 =current of coil C1(Amps) I2 = current of coil C2(Amps) rxx = distance between currents being calculated(meters) b = length of coil section C2 working on C1(vector) r = distance between current section being calculated and I2 now if you need to calculate the net force of coil C1 on coil C2, you would apply the same equations to each coil sides respectively. But you'll find the net force acting on C1 is greater than the force acting on C2. -DandelionTheory
  13. i need some help answering this question about calculating force on a portion of a coil. got it from here: if lines DK < KC, lines AJ < JB, lines AJ = DK,and lines AD = BC i can calculate the magnetic field from each side at point P with: B = [μ0I/4π(L/2)] [Sinα +Sinβ] (the same α, L and β variables used in the video) how do i calculate the force on coil C1 line JK if both coils magnetic fields come out of the screen? (picture, not video) -DT
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