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About DandelionTheory

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  1. The engine applies constant gravity to all rigid bodies and an opposing force is applied to the structure at certain conditions. It's in the script attached to the "Pin" object (center of gravity). What I mean by "defying gravity" is, the center of mass has enough momentum to counter the constant acceleration of gravity. Yes it moves backwards and forwards with less than 570 force units, but with more than 575 force units it gains momentum in the Positive Y direction. The name is supposed to be satire. i messed up, i see a divide, false alarm
  2. If I plug values into a physics engine, would that be enough of a proof of concept? https://forum.unity.com/threads/defying-gravity.858343/ I made something in the unity3D physics engine that shifts it's center of gravity in the positive Y axis at around 600 force units. It would require downloading the engine (version 4.7) and the project file from the forum I posted to view on your computer, of which is asking alot. Its the best I got, I hope it's enough. -DandelionTheory
  3. Pardon, I assumed stating the center of gravity was enough. I assume force applied 90° from center of mass is transformed to torque through the center into the page. I needed clarification as to when torque was an applicable calculation. Do I substitute L with 2πr^2 for each loop? Thank you. Is there torque about C due to the force on I2?
  4. Right, he answered part of it. I asked about a specific part to which no variables were stated in the answer. Unit length of what? How does it work into my problem? I'm not going to applause an answer if its incomplete.
  5. https://www.school-for-champions.com/science/force_torque.htm#.XnmML5BlA0N
  6. Thank you. Just to confirm, The force on I1 is F=qE+I1xB2sin theta, where B2 is measured at I1 The force on I2 is F=I1LxBsin theta, where B1 is measured at I2 and theta is the angle between the current and interacting magnetic field. Is my understanding correct? I've seen videos where torque was described as rF=Mar, but you described it as F/r, I am confused. Please set me straight.
  7. (Given) if the: Center of gravity C is in the center of current I1 Blue X represents the direction of current I1, which is into the page Red box represents an iron core Green arrows represent magnetic fields due to I1 & I2 respectively Black circle next to I2 represents a wire loop with center axis perpendicular to current I1 Black circle next to I1 represents a cross section of wire in which I1 passes through the XY plane, the current loop I1 is not represented intentionally to ask this specific question. B1 designates the magnetic field due to I1 B2 designates the magnetic field within the iron core due to I2 Assuming the force on I1 is a Lorentz force calculation with a small air gap between I1 and B2, (Question) how would one calculate the force and translation experienced by B2&I2, and is that force translated to net motion perpendicular to C(on the xy plane)or torque about the center of gravity? "Debate the topic, not the language" -SomeoneSmarterThanMe
  8. Yup I'm done. You've officially killed my drive for physics. Thanks for the encouragement Go fuck yourselves.
  9. Can you point me in the right direction or is "it's just complicated you wouldn't understand" a viable answer I should be content with? I'm not. I assume charges repel each other. I assume rotational torque is applied to the axle of a wheel at 90°. I assume charges can repel each other and apply force to a wheel at 90°.
  10. So force applied to 2 wheels at 90° throws off their center of gravity? You said that's impossible, survey says! "No" So why can't you jump off a wheel by applying force at 90°?
  11. If you were in space between the bucket ends of 2 water wheels, close enough to put your feet on, and ran up both of them with downward force only, you would pop up above them and they would spin in place. If you jumped but a tether attached to both wheels axis pulled you back, the opposite force would pull them to you. Just run up the wheels again.
  12. What if things rotate? Right, you would need a linear component to transfer momentum to, like circle E. Like letting go of a mary go round.
  13. Let's tackle this with ideas first. You're implying rotation and linear momentum cannot be transformed from one to the other? That's ridiculous.
  14. Can you show me how I would calculate that balance? I'm having trouble finding the right documentation for linear motion on a wheel axle due to different angular torques applied. If you know of anything I would appreciate it. Because I want to know why you came to the conclusion. I just realized I did the statement wrong. Apologies. I'm making a correction to this statement: Should say: Circle E is forced up half the cycle, while no force is experienced by circle E while circle A & B are parallel to line FG. As circles A & B pass line FG to the closest point of rotation to circle E, if circle E were to be manually switched to it's opposite charge it would once again experience force for the remainder of the cycle while circles A, C, B & D rotate
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