Spyman

This thread have some great advices: http://www.scienceforums.net/topic/54213-nicotine-tapering-method-to-quit-smoking/

I am sorry, but I can't give you any guarantees about your inverter's output, as i don't have any experience with it, however at my work we have lots of computers running on UPS power supplies to protect them from surges and brownouts from our mains. The charger in the UPS could still be connected to the battery even if the UPS is disconnected from the mains, if it is powerless it should not interfere with another charger, but I need to stress that I can't give any guarantee that connecting a second charger won't harm the UPS or the charger. EDIT: If you are going to try this you need to know that a 12 volts 20 amperes charger will only give 12 x 20 = 240 Watts. If you need 300 Watts output then you need a 25 ampere charger and for 600 watts you need a 50 amperes charger. The battery can cover for a larger need for a short duration but if you want to run constantly then the charger need to be greater than your average consumption. Also the inverter have losses so you need to make sure you have a little extra for this too.

If it is a good inverter then it will not destroy your appliances, and if you already have used it for longer durations without mishaps then I don't think it will be more harmful for your equipment if you were to always use it. The problem with another separate charger is when your mains voltage goes above 170 VAC and the UPS also tries to charge the battery, then there will be two chargers competing on the battery voltage and if they don't work together one or both could be destroyed. If the UPS allows a separate charger and you can find a charger that is compatible with it, then it would probably work.

If you want 220 VAC independent of input voltage then you need some kind of inverter. It have to automatically switch from bypass mode when voltage drops to low and start to supply 220 VAC but still be able to continue charging its battery or capacitors in the intermediate circuit with the low input voltage. (Or if you have very long periods with low voltage then you could have a separate charger and inverter and manually turn them on when they are needed.) Much like your UPS but with an intermediate mode where it supplies enhanced power while still charging the battery from the low input voltage. I have no idea of what models and brands there are on the market that can fulfill your needs, but I am sure a local salesman or the second repairman can help you. (If your voltage normally is so low, then there should be lots of different equipment to choose from.)

There is nothing in that device that will make it able to output exactly 220 VAC. It has three stages, adding a low voltage to input, removing a low voltage from input or forwarding input to output. (Also if the input voltage goes way too low the relays will not be able to activate and it can only forward input to output.) It is likely that you have changed the settings for the two potentiometers such that the two relays no longer can activate as they should. Finding the correct settings for the potentiometers without access to adjustable voltage is not going to be easy and since it requires adjusting with live voltage it will be dangerous. My recommendation is that you take it back to the second repairman and let him redo the adjustment and check the unit again. But even with the settings back as they were intended, the unit will not be able to give 220 VAC from 90 VAC. However when your main voltage is above 180 VAC it can stabilize the voltage and ease the strain on your UPS.

Yes, but my view of a flat Earth is the old fashioned with a fixed Earth and an absolute up and down. If they believe that the flat Earth is like a spinning coin orbiting the Sun and has the Moon as an orbiting companion, then they have already come very far from the old model and the final step to a spherical Earth is small.

Thank you for clearing up the misunderstanding.

The front implies that the device can handle 90 to 250 VAC but it does not specifically guarantee that it will output 220 VAC from 90 VAC. From the image it looks more like there is one indicator light that shows that the input is between this level and one indicator that shows when the output is around 220 VAC. Did you take note on how they where located so that you can return them to the position the second repairman placed them in? How do you know it gives power if the electric iron didn't work? Good. What voltage did they show with the electric iron connected? Can you get a light bulb for testing? It is not as good as a voltmeter, but satisfies to give a rough voltage indication and is not expensive. You also missed out to reply to this important question:

During a partial lunar eclipse we can look up and see the curved shadow of Earth on the Moon: A lunar eclipse occurs in two regions, an outer penumbral shadow where the sunlight is dimmed, and an inner umbral shadow, where much dimmer sunlight only exists by refraction through the Earth's atmosphere, leaving a red color. This can be seen in different exposures of a partial lunar eclipse, for example here with exposures of 1/80, 2/5, and 2 seconds. https://en.wikipedia.org/wiki/Lunar_eclipse Maybe you and your friend can see it yourselves on September 28: A total lunar eclipse will take place on September 28, 2015. The eclipse will be visible over Europe, the Middle East, Africa, and the Americas. View of Earth from Moon The following simulation shows the approximate appearance of the Moon passing through the earth's shadow. https://en.wikipedia.org/wiki/September_2015_lunar_eclipse

Well, speaking for myself, I can personally confirm that gravity exists and see with my own eyes how the Moon is shaped, logically it is gravity that holds all the parts of a natural celestial object together, thus the Earth has to be round too.

If he believes in a world wide conspiracy that hides the truth of a flat Earth, then why do you think he would trust personal photos from strangers on the internet?

This thread is getting slightly out of hand, first we need to understand and solve the issue with the stabilizer before we can move on to other equipment the OP want to connect to it. ----- I have only looked at a few of the photos since I have serious trouble with dropbox from work and this seems to be a very simple device only able to stabilize voltage above around 180-190 VAC and up to 250-260 VAC with a maximum current of 10 amperes. There is no pulse wide modulation, switching nor capacitor filtering in this unit. I guess it has a low voltage transformer that probably has an output of around 24 to 30 VAC which then is put in series, (by the two relays), with the line supply, such that it either increase or decrease the output voltage. The two potentiometers is there to adjust the voltage limits when one of the relays should activate to alter the output. So for an example if the input goes below 200 VAC, then one relay change state such that the transformator output adds to the line voltage which would then give an output of 200 + 24 = 224 VAC and if the input goes above 250 VAC, then the other relay change state such that the transformator voltage lowers the output voltage to 250 - 24 = 226 VAC. Of course the output voltage of the transformer would also decrease or increase with the same amount as the input voltage changes, which means that around 180 VAC the output would only be around 200 VAC. ----- The second repairman seemed to know what he was talking about and doing, and since the unit apparently worked when he tested it, within the voltage limits he concluded the stabilizer only can handle, have you changed anything or did anything happen afterwards? When the stabilizer is deemed to not be working, is it not working above 180 VAC input or not working below 180 VAC input? The stabilizer seem to have two meters on the front, what do they measure, input, output, voltage or amperes? What does the two meters on the stabilizer show with and without the 1000 watts electric iron connected? What does the two meters on the stabilizer show with and without a standard light bulb connected? Is the light bulb dark, glowing or shining?

That would depend on the potential of the circuit and the charge of the capacitor at the end of the wires. If the battery is connected long enough for the potential to even out and if the battery later is reconnected with the polarity switched, then L3 will light up. For the second circuit the capacitor is almost entirely removed and to small to light up L3.

Well yes, sure they could probably compete or even be considered superior in scientific knowledge and they could likely teach the caveman some useful stuff. But that was not the question, the question is if they can be trusted on everything they claim knowledge of. Should we take their word for it and accept whatever thay say without questioning it, just because they step out from a fancy ride? You seem to argue that anyone with "a functional interstellar spaceship" is all-knowing, infallible and only wants the best for us. I think 'they' could be wrong about many things, make mistakes and even straight out lie to us if it would benefit them.

A caveman would probably learn a lot from listening to a former Resident Expert at SFN, but how much not relevant crap could he be fed if he would meet a random proponent from our Speculations section?

Sounds suspiciously like a trap to me...

But this is a discussion forum, in PMs we can't read and benefit from your arguments.

Extra isolation from the electricity is always good but if you insulate the coil and metal core they can get burning hot. A better description of what you are building and maybe a picture of the device can probably spur more helpful ideas.

Theoretical physicist Kip Thorne was a scientific consultant for the film to ensure the depictions of wormholes and relativity were as accurate as possible. "For the depictions of the wormholes and the black hole," he said, "we discussed how to go about it, and then I worked on the equations that would enable tracing of light rays as they traveled through a wormhole or around a black hole-so what you see is based on Einstein's general relativity equations." https://en.wikipedia.org/wiki/Interstellar_(film)#Scientific_accuracy

Yes, as long as there is a current flowing through the lamps they will shine but if the circuit is open at L2 then they will fade when the potential equalizes. The wires are not independent of each other, if you completely remove one wire then you also completely change the setup. The two wires are like the opposite sides of a huge capacitor or many small in parallel along the wire length. Place a symbolic capacitor at the end of the wires, in parallel with L2, to see how this affects the circuit. During the creation of the circuit with L2 in place and the switch open, both sides of the capacitor will be charged with the negative side of the battery and when the switch is closed the battery will start to charge the positive side of the capacitor with its voltage. Removing L2 will not stop the charging but without L2 there will be no current when the charging has finished.

Yes, but I can't claim expertise, so I might still be wrong about it. There should still be a current until the potential has been fully equalized to battery voltage to the furthest end of both wires. (The wires are like a huge capacitor and they will require lots of movement of electrons to even out the charges.) Then the potential is already at battery voltage at the switch when it gets closed, so L1 lights up immediately. (L3 would light up as before, but L2 would now also be closer to the switch and lights up close to L3.) The electron shells are not physically touching each others, since they are repulsed by their charges like you agree to, but that also means that an outside higher voltage can push them closer together in one end for a short moment before the potential has evened out through this hypothetical straw of yours. The electrons must start to move in one end before they all can move forward together. If you have a light year long iron bar and hit it at one end with a hammer it won't move immidiately at the other end, there will be a pressure wave moving through the iron bar at the speed of sound in iron that eventually push the other end forward. Air is not a good conductor but with high enough voltage you can force a current through it, lightning does this every day on Earth.

Thank you, that was what I meant with my edit. And if the track length would increase proportionally with the numbers of runners then the distance when a runner would feel lonely would be constant. So the problem doesn't get more difficult with increasing or decreasing distance, it's the increasing number of runners covering parts of the track that makes the problem more complicated and harder to prove. If the distance wouldn't be proportionally decreasing with increasing numbers of runners then two runners would never feel lonely.

"Consider k runners on a circular track of unit length. At t = 0, all runners are at the same position and start to run; the runners' speeds are pairwise distinct. A runner is said to be lonely at time t if he is at a distance of at least 1/k from every other runner at time t. The lonely runner conjecture states that each runner is lonely at some time." http://en.wikipedia.org/wiki/Lonely_runner_conjecture#The_conjecture Doesn't a larger k mean more runners on a longer track? (EDIT: Or that the distance between them is allowed to be shorter with more runners.)

If you plan to make a big and deep hole then you also need to think about the safety around it. Children or animals can fall down and get hurt and trapped or drown when filled with water.

Yes, I made an analogy to explain my thinking, but I never claimed it to hold in every case you can think of. I think there will be two potential surges travelling outward in both cables, from both sides of the battery when the switch gets closed. If we place three lamps in the circuit, the first after a half light years, the second at the furthest point a light years away, and the last at a half light years from the battery in the other wire at the opposite side as the first lamp. Like this: ┌──────────────────────────(L1)───────────────────────────┐ \ S │ + │ [B] (L2) - │ I │ └──────────────────────────(L3)───────────────────────────┘ Then both the lamp (L1) and the lamp (L3) will light up at the same time, around half the time before the lamp (L2) will light up.