Umm

E = (Iw2)/2. So where I = 0.00098655969 and
where w = 230.2666 then E = 26.155 @ 2200 rpm

Where w = 157 then E = 12.1588 @ 1500 rpm

And where w =  104.6666 then E = 5.4039 @ 1000 rpm

Need more data now.


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Farmhand,

I am glad that you are having fun.  As a reminder, anything that you can do to reduce the friction and/or cogging that results in a faster RPM after the weight falls is going to reduce your calculated moment of inertia.  That's a good thing and should give you more accurate results.  As long as it's passive and does not add any active power or energy to the spinning rotor you are fine.

About your first round of results:

For your method I get the figure   I = 0.00098655969
And by calculation I get the figure I = 0.00091038461

Note that these results are "satisfying" in the sense that the dropping-weight method shows a larger MOI, and we can suspect that unwanted friction might be the main cause of the higher MOI measurement.  So it's reasonable and could be expected that the dropping-weight MOI measurement should be larger than the other method.  If it was the other way around and the dropping-weight MOI was smaller, then the first round of measurements would not "smell right" which would merit further investigation.

MileHigh

Farmhand,

Apology accepted and it's water under the bridge.  I am not perfect and have done a few outbursts myself.  Just for your knowledge, my real interest these days is to debate the "professional" free energy propositions.  So I debated with John Rohner of Inteligentry infamy on PESN as an example.  I could tell that he was full of crap, and was basically a criminal.  Other contribtors on PESN like Mark E. could see through him just like me.   (If you want to observe someone that really knows his stuff in "action," try reading Mark E. on PESN.  The guy is brilliant.)  About thee months ago the FBI did a raid on Inteligentry for fraudulent solicitation for the selling of shares in his company in violation of US Securities and Exchange Commission rules.  The FBI report also mentioned that he was making false claims about his alleged "PAPP engine."

So that was fun and I got a kick out of that.  I view that as doing some good for the community at large.  These types of "professional" free energy propositions get some coverage on OU from time to time and sometimes the real players will post here.  It's fun to get into a debate with the actual people that are in a high-profile professional looking proposition.  Sometimes I can spot a fake quite quickly because it becomes apparent that they are bluffing and don't know what they are talking about.  This is especially true when talking about electronics because of my background.  It's important to state that I am no genius in electronics by any means.  I was never a big analog electronics fan.  However, for most of the analog electronics discussions you see around here I can pretty much grasp what is going on.  I am not qualified at all to do serious "real" analog design work.

As another high-profile example, on the "Delayed Lenz" thread I looked up Magnacoaster and just discovered that the web site was revamped into a medical quackery web site.  Richard Willis of Magnacoaster is another person that is a fraud.  To bad he didn't dare come here himself to drum up business, because myself and others would have justifiably made mincemeat out of him.  I can tell that he is bluffing when he talks about his electronics circuits.  The Magnacoaster systems have apparently vanished and now he is Doctor Quacko and his Magnetic Belts Inc.

With respect to the goings on here OU, I tend to get lightly involved in the occasional thread for fun.  I don't get deeply involved simply because I have read and commented on many threads in the past so the novelty wore off a long time ago.  So I am glad you are having fun and learning as you go along.  I hope that your neck problem goes away or is at least bearable when it gets bad.  I have had two back operations from a crushed disk and I know how bad it is when there are problems with your nerve system.

I wrote assembler language programs for the 8088 and the 6502 back in the day.  I know nowadays that you program microcontrollers with a high-level language.  There is a kind of "nerd thrill" when you write in assembler because you have full control and have "taken over."  Assembler also runs blazingly fast.

So carry on and have fun!

For even more fun, you might want to look up "The CMOS Cookbook" by Don Lancaster.  That book excited and spawned a whole generation of electronics hobbyists way back when.  You can probably say that it's part of the "Nerd Collective Unconscious."  lol

MileHigh

Tinsel I'm ordering the Arduino Eleven you suggested as a good board http://www.freetronics.com/collections/arduino/products/eleven#.UbBap_nPVqA
And I was wondering if you though it a good idea to get the 16 x 2 LCD Shield for Arduino. http://www.freetronics.com/collections/shields/products/lcd-keypad-shield#.UbBjMPnPVqA

I kinda like the idea of the relay shield as well maybe for motor circuit input power on/off switching or other control switches, battery swapping maybe, things like that. I'll check that out and buy it later maybe.

Couldn't resist, so I got he LCD shield as well. Thanks for the tip on the board it looks very good for what I want to play around with.

Cheers 

I was wondering if someone could tell me if I would really need a pull up resistor with this hall sensor circuit, if I modify the existing optical sensor circuit design and use a CMOS inverter chip with the hall sensor so I get both non inverted and inverted outputs, would the output (of the Hall sensor) not just go low when the magnetic field exceeds the threshold and then go high when the field is not exceeding the threshold ? Why the need for the pull up resistor ? The data sheet says I need a suitable pull up resistor.

Also says this.

http://pdf1.alldatasheet.com/datasheet-pdf/view/55092/ALLEGRO/A3144.html

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Oh MileHigh, I've had the CMOS cookbook for some time now.  I occasionally recommend it myself.

I've made some CMOS logic chip oscillators that can reach stable frequencies of over 5 mHz with just a cheap CD4001, the better inverting chips can do better, CD 4049 I think.
It's all layed out in the cookbook, it's an awesome resource, SeaMonkey recommended it to me a year or so ago.

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The Allegro sensor that you have chosen doesn't really have an "output", unless you provide it with an "input". It is an open-collector-output device. You treat its Pin 3-Ground output like the Collector-Emitter portion of a NPN bipolar transistor that can sink 25 mA. So you just treat it like a normal bipolar transistor in your circuit, with a suitable pullup chosen to keep the current below 25 mA when the transistor turns ON from the magnet passage. The value of the pullup will depend on the voltage at the positive rail and the downstream device you are switching.

In other words you can't get a logic "high" signal value out of it unless you provide it with a "high" voltage value to switch.

(ETA This makes it really easy to interface with microprocessor boards like Arduino. The same sensor circuit can be used with photodiodes, phototransistors, photoresistors or just about any other thing that acts like a switch when stimulated.)

Allegro also makes 2-wire Hall sensors that act just like a reed switch in a circuit (except without all the problems with reeds).

Umm

E = (Iw2)/2. So where I = 0.00098655969 and
where w = 230.2666 then E = 26.155 @ 2200 rpm

Where w = 157 then E = 12.1588 @ 1500 rpm

And where w =  104.6666 then E = 5.4039 @ 1000 rpm

Need more data now.


...

Heh... not to be too harsh or anything.. but _significant digits_ only, please! You cannot expect me to believe that you have measured your current or the values that go into calculating it to the degree of precision that a number like 0.00098655969 implies. You are here saying that the actual value of the current is 0.00098655969 amps, and NOT 0.00098655968 amps or  0.00098655970 amps.... or any other value. So I say to you that your current value is wrong! I could not say this if you had cited your current value as 1 milliamp, though. You would then be right! because the 0.001 value represents the degree of accuracy of your measurement. If you had given it as 0.001000000, you would once again be wrong.

The answer of any computation cannot legitimately include more digits of precision than the _least_ precise value that goes into the calculation. Those extra digits are just fantasy; the odds of them being exactly correct are smaller and smaller the more of them there are.

Umm What current ? That's not a current calculation. That is the energy in the rotor. I did the calculations and those are the figures I got as the result. I don't understand what else I could do except leave off some numbers, which would surely give a less accurate result, that's all it can do. Can you explain to me how I could do the calculations without getting decimal points ?

Sorry, I thought I was looking at a current calculation. The "I" confused me. But the point still stands. You calculated that MoI figure somehow. From measurements and weights, and somewhere in there you made a measurement that is only precise, to say 3 digits. Like 0.756 grams or something, or 0.625 millimeters. This means that any calculation you do with that measurement can't be more precise than that. And I used the rounding to 0.001 in your example as the extreme case. You probably can round to three sig digs. Please look up "significant digits" or figures in the Wiki.
http://en.wikipedia.org/wiki/Significant_figures

The hall sensor I realize it's an open collector device, and it will be supplied 5 volts to the supply pin of course, (as the drawing shows) I couldn't expect it to work without any input. The CMOS chip has a high impedance so I don't see how much current will flow to require the pull up resistor.

I think you are misunderstanding how open-collector devices work. The power supply pin to the Hall chip does not produce a "high" signal at the open collector output, it only supplies the chip's sophisticated electronics inside. The presence or absence of a triggering event closes or opens the collector-emitter channel of the sensor's output transistor. That is all. Just like the block diagram shows. The CMOS chip that you are driving will not consume much current at its high impedance inputs, and so you can raise the value of the pullup to limit current flow thru the Hall open-collector output transistor to the minimum value consistent with triggering. When the sensor is not sensing, the collector-emitter channel will be open, and so the voltage at the junction of the collector and the pullup resistor will be at the positive rail voltage (logic HI). When the sensor triggers, the channel closes and so the voltage at the collector-resistor junction falls to near the negative rail voltage (zero, logic LO). It is a bipolar transistor in there, wired as the block diagram shows! You have to provide a voltage for the transistor to switch, there is no internal connection to the collector. Open collector. Just as if it were a photodiode or phototransistor with no base lead at all.
The emitter pin is of course shared by the bipolar transistor output and the chip's power supply ground.

I don't get the replies, but that's OK because I can work on alone. No problem.

Don't worry about being Harsh, as long as it is said when I am misunderstood I don't mind at all.

I mean to say you provided me with the formula, can you show me then how I should have calculated it without all the decimal points ?

I don't understand what it is that I didn't show that was so confusing ? The formula is there. Are the calculations correct or not ? Please say so either way.

The calculations are "correct" but wrong, and hopefully you will understand after you read up on "significant digits". Just because your calculator has all those digits doesn't mean they are always meaningful.

You also spoke of a reed which I don't use, so I can only assume you are not taking much notice of what I post.

I spoke of Allegro 2-wire Hall sensors, which function like reed switches in that they only need two wires, instead of three, to produce a signal. Your schematic would be usable directly with one of these sensors, probably. I can only assume that you are looking deliberately for ways to misunderstand what I'm telling you. I thought you had enough electronic knowledge to know right away how a bipolar transistor is used in a circuit, and how to read Allegro's data sheet for your sensor. Sorry, my mistake.

If you don't want to help then fair enough, just say so and I'll stop wasting time asking questions here and go elsewhere. I won't be offended, you are busy with debunking, I know.

How do you possibly get "you don't want to help" out of my trying to explain to you how to use your sensor, in direct response to your question?
That is going to take some explaining for me to understand.

MileHigh has bowed out and said he no longer wants to help, which is fair enough. He's not here to help us only to debunk, that is clear now.

I'll just use a solderless board and spend time working things out for myself.

Cheers

.

Have fun.

I'm just playing with youse, I know how the open collector output works, I've been using SG 3524 and TL494 chips for some time now a couple of years. I was just demonstrating how frustrating some people can be.

I keep being misunderstood. It peeves me.

I asked you MileHigh several times for my calculation results to be checked, no response on that.

Never mind. I'm Busy.

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