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  • Ufopolitics
    replied
    @Citfta/Bistander...

    @Citfta/Bistander...

    By the way, you are so obvious, it really pops out like some sort of huge Tongue...

    My last contact with your second identity, Citfta was more than a week ago (10-06-15), and now that original post from Citfta is gone, deleted...am glad I quoted it...

    However, yesterday, after my discussion with you as Bistander here...you logged in as Citfta and made a post against me all the sudden... in the Thread Citfta has created...that was so stupid from your end.

    Then you log in again as Bistander to write the stupid comment below, so stupidly trying to make up wordings and expressions denoting a different "accent" none of your two characters uses...:

    Originally posted by bistander View Post
    (to Citfta...)...I wondered what happened to you. I'll be sorry to see you leave. Did you notice on his million veiw thread he is claiming you and I are one in the same? Oh well. Ya gotta do what you gotta do. Good luck and God bless.

    bi
    You are so ridiculous.

    So now that your dual identities have been blown out, Burnt, exposed yesterday by me...then now you have decided to kill Citfta, sending it to another Forum...?


    That was so ridiculous, so stupid, and so obvious...your personality is just like that wall across the road...

    Do you really believe people here are stupid?

    Go somewhere else to post, not here...neither Citfta or Bistander are welcome here...so, if you want to keep addressing me, try another identity.
    Last edited by Ufopolitics; 10-15-2015, 06:18 PM.

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  • Ufopolitics
    replied
    The final conclusion here is that you have previously ONLY calculated the On Time, but not the frequency or REPETITION it takes place during the TIME the rotational speed develops.

    So, based on an incomplete result you drew all kind of wrong conclusions.

    I am not wasting any more time here.


    write whatever you please...be my guest.

    Leave a comment:


  • bistander
    replied
    Originally posted by Ufopolitics View Post
    I really can not tell if you are so retarded for real, or is it your obsession and stubbornness to demonstrate your point, proving me wrong.

    So, let me restart/refresh from your previous post above with your results.

    Your results, which only means "time of contact" (as you have written above)...trying to translate it to you ...:

    It only means the time the brush establishes contact with commutator segment...which is perfect, brilliant, excellent...However, Those results does NOT reflect how many times that "time of contact" REPEATS related to the Main Frame Time from Shaft, and of course commutator which is mounted thereto.

    Based on normal and simple logic, no one with a healthy and calmed mind... could think that 6.1 or 9.5 milliseconds where one (1) Second contains One Thousand (1000) Milliseconds...and one (1) minute contains 60,000 milliseconds...is going to represents the repetition value on a shaft and commutator spinning at 1000 Revolutions Per Minute or 16.666 Revolutions Per Second.

    So after you did your excellent calculation about "time of contact" between brush-commutator segment ...we need to know now how many times that value (time of contact) repeats within one minute (60 seconds) AND also for one Second of rotation from main shaft...:

    First let's convert your results back from milliseconds to seconds:

    Case 1- For the 1" radius Commutator (9.5 mS) = 0.0095 seconds

    Case 2- For the 4" radius Commutator (6.1 mS) = 0.0061 seconds

    For One (1) Second of Rotation...:

    Case 1 (r=1"comm) = (1.0 Sec/0.0095 Sec)=105.26

    Case 2 (r=4" comm) = (1.0 Sec/0.0061 Sec)=163.93

    So, on the four inches radius commutator, the time of contact REPEATS around Sixty (60) times MORE than for the Case 1, or one inch radius commutator...and that is ONLY for One(1) second of rotation.

    Now, One Minute has Sixty Seconds right?...so then:

    For One (1) Minute of Rotation:

    Case 1- 1" Radius= 1 minute= 60 Sec/0.0095 Sec= 6,315.79

    Case 2- 4" Radius= 1 minute= 60 Sec/0.0061 Sec= 9,836.06

    So, in one minute the 4" commutator segment establishes contact with brush roughly around 3520 more than the one inch commutator...

    Now that is a big time difference between those two different diameter commutators...considering some motor applications stay operating for very long periods of time.

    Not that "unnoticeable" now right?

    Now I am resting this case here Bistander, it is completely up to you to keep refuting you are completely wrong...but you will be doing it by yourself


    To me it is the end of this non sense argument.


    Ufopolitics
    Hi Ufopolotics,

    Great example of what I call bad math. You also have a habit of bad English when it comes to singular and plural. So right now it is unclear what you are attempting to calculate. For a 1 second duration, you come to the conclusion of 105.26 and 163.93. What are the units for those numbers? What do they represent? The same question for a 1 minute duration. What are the units for 6,315.79 and 9836.06?

    Those numbers can't be the number of times a comm segment encounters a particular brush. It is obvious that is defined by the 1000RPM, so it is 1000 times per minute or 16.7 times per second, for any diameter commutator.

    From post #7917 I calculated that for a 1" radius and a 4" radius commutator the segment time of engagement with a brush is 9.5mS and 6.1mS respectively for each revolution at a 1000RPM. So, if you want to know the total time the brush contacts that segment per minute, you just multiply the rotational speed times the contact time per revolution.

    9.5mS per revolution times 1000RPM = 9.5seconds per minute or 15.8%.

    6.1mS per revolution times 1000RPM = 6.1seconds per minute or 10.1%.

    9.5mS per revolution times 16.7 revolutions per second = 158mS per second or 15.8%.

    6.1mS per revolution times 16.7 revolutions per second = 101mS per second or 10.1%.

    These percentages represent the fraction of time that the particular commutator segment touches a certain brush while rotating at 1000RPM. These percentages would hold for any speed above zero with the specified size commutator and brush.

    *The difference of contact time between the 4" and 1" commutators would be the fraction of the percentages. 10.1% divided by 15.8% = 64%. So the contact time for the larger comm segment with that brush is 64% of the time of the contact using the smaller com. The same percentage you get when you divide 6.1mS by 9.5mS.

    Seems simple to me. But maybe I am just too old fashion to comprehend your math. If so, please continue your explanation so I might learn.

    Regards,

    bi
    Last edited by bistander; 10-15-2015, 04:51 PM. Reason: addition *

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  • Ufopolitics
    replied
    Refreshing...again.

    Originally posted by bistander View Post
    So let's use the example from before. Replace the small comm (1"radius) with a larger comm (4" radius) and leave the smaller brush which had a contact of 30 on the smaller comm or an arc length of 0.523". Use 1000RPM for both cases.

    The smaller comm surface speed is 104.67in/sec. The comm segment actual surface distance would be 0.473" when a .050" mica undercut is considered. The total time of contact between the brush and one comm segment = (0.523"+0.473")/104.67in/sec = 9.5mS.

    The larger comm surface speed is 418.67in/sec. The comm segment actual surface distance would be 2.045" using the same .050" undercut. The total time of contact between the brush and one comm segment = (0.523"+2.045")/418.67in/sec = 6.1mS.

    The contact duration is shorter on the larger commutator.

    This would have negligible effect on torque and speed of the motor, probably unnoticeable.

    bi

    I really can not tell if you are so retarded for real, or is it your obsession and stubbornness to demonstrate your point, proving me wrong.

    So, let me restart/refresh from your previous post above with your results.

    Your results, which only means "time of contact" (as you have written above)...trying to translate it to you ...:

    It only means the time the brush establishes contact with commutator segment...which is perfect, brilliant, excellent...However, Those results does NOT reflect how many times that "time of contact" REPEATS related to the Main Frame Time from Shaft, and of course commutator which is mounted thereto.

    Based on normal and simple logic, no one with a healthy and calmed mind... could think that 6.1 or 9.5 milliseconds where one (1) Second contains One Thousand (1000) Milliseconds...and one (1) minute contains 60,000 milliseconds...is going to represents the repetition value on a shaft and commutator spinning at 1000 Revolutions Per Minute or 16.666 Revolutions Per Second.

    So after you did your excellent calculation about "time of contact" between brush-commutator segment ...we need to know now how many times that value (time of contact) repeats within one minute (60 seconds) AND also for one Second of rotation from main shaft...:

    First let's convert your results back from milliseconds to seconds:

    Case 1- For the 1" radius Commutator (9.5 mS) = 0.0095 seconds

    Case 2- For the 4" radius Commutator (6.1 mS) = 0.0061 seconds

    For One (1) Second of Rotation...:

    Case 1 (r=1"comm) = (1.0 Sec/0.0095 Sec)=105.26

    Case 2 (r=4" comm) = (1.0 Sec/0.0061 Sec)=163.93

    So, on the four inches radius commutator, the time of contact REPEATS around Sixty (60) times MORE than for the Case 1, or one inch radius commutator...and that is ONLY for One(1) second of rotation.

    Now, One Minute has Sixty Seconds right?...so then:

    For One (1) Minute of Rotation:

    Case 1- 1" Radius= 1 minute= 60 Sec/0.0095 Sec= 6,315.79

    Case 2- 4" Radius= 1 minute= 60 Sec/0.0061 Sec= 9,836.06

    So, in one minute the 4" commutator segment establishes contact with brush roughly around 3520 more than the one inch commutator...

    Now that is a big time difference between those two different diameter commutators...considering some motor applications stay operating for very long periods of time.

    Not that "unnoticeable" now right?

    Now I am resting this case here Bistander, it is completely up to you to keep refuting you are completely wrong...but you will be doing it by yourself


    To me it is the end of this non sense argument.


    Ufopolitics
    Last edited by Ufopolitics; 10-15-2015, 03:46 PM.

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  • bistander
    replied
    Originally posted by Ufopolitics View Post
    Nope, you ruined my whole day, now it is my time.

    I do know exactly "how many revolutions are in a RPM"

    Exactly ONE, Your motor was going 1000 RPM's, meaning One thousand revolutions PER ONE MINUTE.

    Your results were given in Milliseconds, and you just wrote previously: each commutator segment will engage a particular brush once per revolution?!

    Your math is so bad, I really can not describe it...and the worst part is that it is the very simple math learned in elementary school!!

    If your totals were given in Milliseconds, then you must convert RPM to RPS, or revolutions per second...then into milliseconds to see how many times each segment will hit each brush during the equivalent "full Minute".

    You do that by dividing the converted total and previous 1000 RPM's in Revolutions per Milliseconds by your previous and great results given in milliseconds...

    Or you could take Milliseconds to Minutes...many fractions there...

    ONLY THEN it will give you exactly how many times each commutator segment will hit a brush per "Minute Revolution".

    Don't try now to avoid this subject, because many thousands of people are watching it...and you are gonna look very bad.
    Hi Ufopolitics,

    O.K. I'll not try to avoid this subject. So first off, let's go back to this question from you:

    Originally posted by Ufopolitics View Post
    ... now which one of the two examples will make contact more times during a full Revolution Per Minute going at One Thousand times?
    I responded as follows:
    Originally posted by bistander View Post
    What? That doesn't make any sense.
    It was a poorly worded question to put it mildly. "During a full Revolution Per Minute"???? What does that mean? A revolution? A minute? How am I supposed to know? So I made the following statement:

    Originally posted by bistander View Post
    Each commutator segment will engage a particular brush once per revolution.
    That statement is correct. And it is correct regardless of the commutator size. So when we take a further look at your original question:

    Originally posted by Ufopolitics View Post
    That's correct, now which one of the two examples will make contact more times during a full Revolution Per Minute going at One Thousand times?

    Answer: Obviously the shorter "Interval" time one...or the larger commutator.
    You say the larger comm will contact the brush more times at 1000RPM. Obviously incorrect. A particular comm segment will contact a certain brush once per revolution regardless of comm size. So the number of contacts per time interval is the same regardless of comm size.

    Originally posted by Ufopolitics View Post
    ONLY THEN it will give you exactly how many times each commutator segment will hit a brush per "Minute Revolution".
    What is a "Minute Revolution"?

    This is an example of bad math. I rest my case.

    Been fun. Go back to Mack's magnet motor. Good luck with that, seriously.

    bi

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  • Ufopolitics
    replied
    Originally posted by bistander View Post
    What? I know how many seconds are in a minute. And I did the conversion. My units are correct.

    It would appear that you have difficulty understanding how many revolutions are in a RPM.

    Just drop it and go do your other stuff. Sorry to have upset you.

    bi
    Nope, you ruined my whole day, now it is my time.

    I do know exactly "how many revolutions are in a RPM"

    Exactly ONE, Your motor was going 1000 RPM's, meaning One thousand revolutions PER ONE MINUTE.

    Your results were given in Milliseconds, and you just wrote previously: each commutator segment will engage a particular brush once per revolution?!

    Your math is so bad, I really can not describe it...and the worst part is that it is the very simple math learned in elementary school!!

    If your totals were given in Milliseconds, then you must convert RPM to RPS, or revolutions per second...then into milliseconds to see how many times each segment will hit each brush during the equivalent "full Minute".

    You do that by dividing the converted total and previous 1000 RPM's in Revolutions per Milliseconds by your previous and great results given in milliseconds...

    Or you could take Milliseconds to Minutes...many fractions there...

    ONLY THEN it will give you exactly how many times each commutator segment will hit a brush per "Minute Revolution".

    Don't try now to avoid this subject, because many thousands of people are watching it...and you are gonna look very bad.
    Last edited by Ufopolitics; 10-13-2015, 08:55 PM.

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  • bistander
    replied
    Originally posted by Ufopolitics View Post
    You ask for this, you started this whole thing by citing my name, and saying I did "Bad Math"...and by getting your nose where no one, absolutely nobody called you and ask for your stupid opinions.

    It don't matter IF what you think works for you is what it is, because it don't work that simple way...you got involved in this argument that Mark and I had from a long time ago...and you obviously were just trying to "capture" something wrong...disregarding the main essence of the whole conversation.

    So now you just shut up an read me well.

    I was talking "Affecting Performance" the diameter of a commutator with same brush...so, it is not as simple as to search for "surface speed" in an online calculator..."online" you are not gonna find what the real answer is, first, because you think you know it all when in reality know absolutely NADA.



    What's the matter?...now you don't even know what your calculations were about?

    NEGATIVE, NOT "ONCE PER REVOLUTION"

    You were calculating based on "Surface speed" in Inch per Second...remember?

    And "RPM" means Revolutions Per MINUTE, NOT PER SECOND.

    So your "Grand Total" of Milliseconds are just NOT applicable DIRECTLY to "once per revolution" based on RPM's which happens to be given in Minutes.

    And How many seconds are in a Minute?

    Come on, don't tell me you did not know that you must convert "Units" to same "Units" for comparison or calculations...in this case "Time Units"

    I am sorry, but ...Who's doing the "Bad Math" now?

    This argument can go much deeper as I stated in my previous post about real contact segments sums, applicable to Symmetrical DC Brushed Motors.
    What? I know how many seconds are in a minute. And I did the conversion. My units are correct.

    It would appear that you have difficulty understanding how many revolutions are in a RPM.

    Just drop it and go do your other stuff. Sorry to have upset you.

    bi

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  • Ufopolitics
    replied
    Originally posted by bistander View Post
    What? That doesn't make any sense. Each commutator segment will engage a particular brush once per revolution.
    You ask for this, you started this whole thing by citing my name, and saying I did "Bad Math"...and by getting your nose where no one, absolutely nobody called you and ask for your stupid opinions.

    It don't matter IF what you think works for you is what it is, because it don't work that simple way...you got involved in this argument that Mark and I had from a long time ago...and you obviously were just trying to "capture" something wrong...disregarding the main essence of the whole conversation.

    So now you just shut up an read me well.

    I was talking "Affecting Performance" the diameter of a commutator with same brush...so, it is not as simple as to search for "surface speed" in an online calculator..."online" you are not gonna find what the real answer is, first, because you think you know it all when in reality know absolutely NADA.

    Each commutator segment will engage a particular brush once per revolution.
    What's the matter?...now you don't even know what your calculations were about?

    NEGATIVE, NOT "ONCE PER REVOLUTION"

    You were calculating based on "Surface speed" in Inch per Second...remember?

    And "RPM" means Revolutions Per MINUTE, NOT PER SECOND.

    So your "Grand Total" of Milliseconds are just NOT applicable DIRECTLY to "once per revolution" based on RPM's which happens to be given in Minutes.

    And How many seconds are in a Minute?

    Come on, don't tell me you did not know that you must convert "Units" to same "Units" for comparison or calculations...in this case "Time Units"

    I am sorry, but ...Who's doing the "Bad Math" now?

    This argument can go much deeper as I stated in my previous post about real contact segments sums, applicable to Symmetrical DC Brushed Motors.
    Last edited by Ufopolitics; 10-13-2015, 08:12 PM.

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  • DadHav
    replied
    Originally posted by Ufopolitics View Post
    Dadhav,

    Obviously you had to jump Huh?...It is time to come and beat the crap out of U.P....So let's go huh?

    Old man, please, stay away from this...as I do respect your person, your work and your white hair.

    Obviously you have nothing else -productive- to do but come around here, basically at this times...

    That is not my case, I am very busy in productive things and hate to be wasting time with this "Member" Bistander/Citfta
    UFO, You'll never change, I wasn't trying to beat the crap out of you, I was only trying to state the obvious and that's the fact that you both have reason to discuss your views on the subject. I just see it as I mention in simple terms for my own satisfaction. I don't give a crap about your argument with Bi. but I do give a crap about your personal insults. It's not in very good taste and it's probably the reason you're almost alone here. Well that and the fact that there's no proof of over unity as you claimed from the beginning of the thread.
    J
    Last edited by DadHav; 10-13-2015, 08:41 PM.

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  • bistander
    replied
    Originally posted by Ufopolitics View Post
    That's correct, now which one of the two examples will make contact more times during a full Revolution Per Minute going at One Thousand times?
    .
    What? That doesn't make any sense. Each commutator segment will engage a particular brush once per revolution.

    Leave a comment:


  • Ufopolitics
    replied
    Why the honor?

    Originally posted by DadHav View Post
    With respect to both UFO and Bi, I've been reading this and just want to see if I'm getting what I should out of it. Didn't UFO state there would be more contact time on a larger diameter brush and commutator than a smaller one? Without math you should have equal contact time providing the brush and the commutator piece have been scaled proportionately. No? The smaller set would have the same RPM and contact time with less surface speed over the contact surfaces. Now UFO said both brushes would be the same size and actually that would change things. Of course if your smaller diameter commutator has to big of a brush you have more problems than just timing. Just saying. I don't think the earlier drawings and explanations matched and I see how an argument can come of it.
    John
    Dadhav,

    Obviously you had to jump Huh?...It is time to come and beat the crap out of U.P....So let's go huh?

    Old man, please, stay away from this...as I do respect your person, your work and your white hair.

    Obviously you have nothing else -productive- to do but come around here, basically at this times...

    That is not my case, I am very busy in productive things and hate to be wasting time with this "Member" Bistander/Citfta

    Leave a comment:


  • Ufopolitics
    replied
    It is NOT that simple.

    Originally posted by bistander View Post
    So let's use the example from before. Replace the small comm (1"radius) with a larger comm (4" radius) and leave the smaller brush which had a contact of 30 on the smaller comm or an arc length of 0.523". Use 1000RPM for both cases.

    The smaller comm surface speed is 104.67in/sec. The comm segment actual surface distance would be 0.473" when a .050" mica undercut is considered. The total time of contact between the brush and one comm segment = (0.523"+0.473")/104.67in/sec = 9.5mS.

    The larger comm surface speed is 418.67in/sec. The comm segment actual surface distance would be 2.045" using the same .050" undercut. The total time of contact between the brush and one comm segment = (0.523"+2.045")/418.67in/sec = 6.1mS.

    The contact duration is shorter on the larger commutator.

    This would have negligible effect on torque and speed of the motor, probably unnoticeable.

    bi
    The contact duration is shorter on the larger commutator.
    That's correct, now which one of the two examples will make contact more times during a full Revolution Per Minute going at One Thousand times?

    Answer: Obviously the shorter "Interval" time one...or the larger commutator.


    But, unfortunately, if we apply this examples you have written to Symmetry (which is exactly what I was referring to previously) both calculations are useless, completely off, just because the actual contacting segments-to energize the coil circuit/angles- would be the sum of all segments between positive and negative brushes for either a two or a four brush system.

    In a two brush system (apart by 180 degrees) the commutator is divided by that brush plane in exactly two parts, where one side would have current flowing opposite to the other side. In a 12 segments that would be at least six (6) per side.

    Still in a four brush system, and even there are two brush planes dividing in four the commutator, still the contacting "area" is not "exactly" one segment, but in the cited case above of twelve (12) segments, it would be -at least- 3 segments in each quarter.

    I am really busy, and do not have the time for whatever BS you are trying to "prove" here.
    Last edited by Ufopolitics; 10-13-2015, 07:42 PM.

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  • bistander
    replied
    Originally posted by DadHav View Post
    Without math you should have equal contact time providing the brush and the commutator piece have been scaled proportionately.
    Well, Yes. That has been obvious to me from the start. And exactly what I have said. Thanks.

    bi

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  • bistander
    replied
    Originally posted by Ufopolitics View Post
    If I replace JUST commutator, not brushes, ...
    In larger diameter than OEM Comm, coils STAY ON LONGER TIME...
    So let's use the example from before. Replace the small comm (1"radius) with a larger comm (4" radius) and leave the smaller brush which had a contact of 30 on the smaller comm or an arc length of 0.523". Use 1000RPM for both cases.

    The smaller comm surface speed is 104.67in/sec. The comm segment actual surface distance would be 0.473" when a .050" mica undercut is considered. The total time of contact between the brush and one comm segment = (0.523"+0.473")/104.67in/sec = 9.5mS.

    The larger comm surface speed is 418.67in/sec. The comm segment actual surface distance would be 2.045" using the same .050" undercut. The total time of contact between the brush and one comm segment = (0.523"+2.045")/418.67in/sec = 6.1mS.

    The contact duration is shorter on the larger commutator.

    This would have negligible effect on torque and speed of the motor, probably unnoticeable.

    bi

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  • DadHav
    replied
    Originally posted by Ufopolitics View Post
    Maybe in your classic physics books did not tell you...you will have to find out by yourself, by experimenting.

    If I replace JUST commutator, not brushes, and put either a smaller or bigger diameter ones, I am affecting the CONTACT TIME to Energize Coils, therefore, performance.

    In larger diameter than OEM Comm, coils STAY ON LONGER TIME every 180 cycle (for symmetry), therefore, magnetic drag is greater related to stators interactions, hence torque is greater, speed is reduced.

    The complete opposite process occurs when smaller diameter commutators are installed.

    The only one who is ridiculous here is you, and your ego thinking you know it all, plus criticizing everyone here with your DOUBLE IDENTITIES.
    With respect to both UFO and Bi, I've been reading this and just want to see if I'm getting what I should out of it. Didn't UFO state there would be more contact time on a larger diameter brush and commutator than a smaller one? Without math you should have equal contact time providing the brush and the commutator piece have been scaled proportionately. No? The smaller set would have the same RPM and contact time with less surface speed over the contact surfaces. Now UFO said both brushes would be the same size and actually that would change things. Of course if your smaller diameter commutator has to big of a brush you have more problems than just timing. Just saying. I don't think the earlier drawings and explanations matched and I see how an argument can come of it.
    John

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