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What's the output of a record needle playing an out-of-speed record


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$begingroup$


I'm very interested in vinyl records and analog music, and the belt of my turntable got loose. Upon such situation it piqued my curiosity, what is the output signal at the end of the arm cartridge wires for a known waveshape if the speed is not the correct one.



Say the record was mean to play sin(wt), a pure sine wave, at 33rpm, then, because of a loosen belt or any other reason, it rotates at a different RPM, how to calculate the changes in such sine wave?



I'm not considering the filters that the needle might apply on the signal, whether it is a low pass, band pass, or high pass nor any other impedances that might alter the signal in any circumstance, just a supposedly ideal needle and cartridge.










share|improve this question











$endgroup$



















    2












    $begingroup$


    I'm very interested in vinyl records and analog music, and the belt of my turntable got loose. Upon such situation it piqued my curiosity, what is the output signal at the end of the arm cartridge wires for a known waveshape if the speed is not the correct one.



    Say the record was mean to play sin(wt), a pure sine wave, at 33rpm, then, because of a loosen belt or any other reason, it rotates at a different RPM, how to calculate the changes in such sine wave?



    I'm not considering the filters that the needle might apply on the signal, whether it is a low pass, band pass, or high pass nor any other impedances that might alter the signal in any circumstance, just a supposedly ideal needle and cartridge.










    share|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      I'm very interested in vinyl records and analog music, and the belt of my turntable got loose. Upon such situation it piqued my curiosity, what is the output signal at the end of the arm cartridge wires for a known waveshape if the speed is not the correct one.



      Say the record was mean to play sin(wt), a pure sine wave, at 33rpm, then, because of a loosen belt or any other reason, it rotates at a different RPM, how to calculate the changes in such sine wave?



      I'm not considering the filters that the needle might apply on the signal, whether it is a low pass, band pass, or high pass nor any other impedances that might alter the signal in any circumstance, just a supposedly ideal needle and cartridge.










      share|improve this question











      $endgroup$




      I'm very interested in vinyl records and analog music, and the belt of my turntable got loose. Upon such situation it piqued my curiosity, what is the output signal at the end of the arm cartridge wires for a known waveshape if the speed is not the correct one.



      Say the record was mean to play sin(wt), a pure sine wave, at 33rpm, then, because of a loosen belt or any other reason, it rotates at a different RPM, how to calculate the changes in such sine wave?



      I'm not considering the filters that the needle might apply on the signal, whether it is a low pass, band pass, or high pass nor any other impedances that might alter the signal in any circumstance, just a supposedly ideal needle and cartridge.







      brushless-dc-motor






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited 1 hour ago









      Dave Tweed

      123k9152266




      123k9152266










      asked 1 hour ago









      Gabriel SantosGabriel Santos

      213




      213






















          4 Answers
          4






          active

          oldest

          votes


















          4












          $begingroup$


          Say the record was mean to play sin(wt), a pure sine wave, at 33rpm, then, because of a loosen belt or any other reason, it rotates at a different RPM, how to calculate the changes in such sine wave?




          The pitch and tempo will change in proportion to the speed change. At 33 RPM it would already be musically flat as the correct speed is 331/3 RPM. A 1 kHz test tone - common on test records - would, at 33 RPM, give off $ frac {33}{33.33} text {kHz} $.



          The sinewave would remain a sinewave but stretched in time and, therefore, a lower pitch.






          share|improve this answer











          $endgroup$





















            3












            $begingroup$

            To really simplify, a record has wiggles in the groove that correspond to the recorded sound pressure. (This ignores stereo, and any compounding, but it answers your question).



            Events are recorded onto that wiggly grove as they happen -- you can think of the groove as a picture of the sound, with the time domain turned into events happening as the needle follows the groove.



            If you play the record slower, all the events happen more slowly -- the singer sings slower and deeper, the orchestra does too, etc. Speeding it up does the opposite -- a normal recording, sped up, sounds like a hyperactive chipmunk.






            share|improve this answer











            $endgroup$









            • 1




              $begingroup$
              I was surprised when I first learned how records work. It's so analog that it's amazing it works at all.
              $endgroup$
              – Toor
              1 hour ago



















            2












            $begingroup$

            Changing the speed of the platter simply affects how fast the groove is moving under the needle, nothing else.



            A sine wave with the time axis compressed or expanded is still a sine wave. In fact, since the groove is a direct mechanical representation of the original complex waveform, you still get the same waveform simply compressed or expanded in time.






            share|improve this answer









            $endgroup$













            • $begingroup$
              Right, agreed, in the ideal case a a sine wave of same amplitude, but with a diferent frequency, right? The point is for a sin(wt), how the change of rotation speed will affect the frequency?
              $endgroup$
              – Gabriel Santos
              1 hour ago










            • $begingroup$
              It's linear -- double the speed means double the frequency. That's what compressing the time axis means.
              $endgroup$
              – Dave Tweed
              1 hour ago



















            0












            $begingroup$

            Grooves are cut with frequency correction according to RIAA equalization. Playing the record off with wrong speed increases all frequencies by the same factor (corresponding to a shift left/right on the frequency axis of the doubly logarithmic transfer function diagram). Since the frequency correction is not a straight line, this does not just result in a frequency shift but also in an uneven frequency response due to recording and replaying correction no longer being proper inverses.



            In addition, the equalization is done in order to reduce excessive signal amplitudes on stylus and pickup. Counteracting this by wrong speed may lead to either excessive amplitudes (electrical or mechanical) or too low signals overlaid with a relatively higher noise floor.





            share









            $endgroup$














              Your Answer





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              4 Answers
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              4 Answers
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              4












              $begingroup$


              Say the record was mean to play sin(wt), a pure sine wave, at 33rpm, then, because of a loosen belt or any other reason, it rotates at a different RPM, how to calculate the changes in such sine wave?




              The pitch and tempo will change in proportion to the speed change. At 33 RPM it would already be musically flat as the correct speed is 331/3 RPM. A 1 kHz test tone - common on test records - would, at 33 RPM, give off $ frac {33}{33.33} text {kHz} $.



              The sinewave would remain a sinewave but stretched in time and, therefore, a lower pitch.






              share|improve this answer











              $endgroup$


















                4












                $begingroup$


                Say the record was mean to play sin(wt), a pure sine wave, at 33rpm, then, because of a loosen belt or any other reason, it rotates at a different RPM, how to calculate the changes in such sine wave?




                The pitch and tempo will change in proportion to the speed change. At 33 RPM it would already be musically flat as the correct speed is 331/3 RPM. A 1 kHz test tone - common on test records - would, at 33 RPM, give off $ frac {33}{33.33} text {kHz} $.



                The sinewave would remain a sinewave but stretched in time and, therefore, a lower pitch.






                share|improve this answer











                $endgroup$
















                  4












                  4








                  4





                  $begingroup$


                  Say the record was mean to play sin(wt), a pure sine wave, at 33rpm, then, because of a loosen belt or any other reason, it rotates at a different RPM, how to calculate the changes in such sine wave?




                  The pitch and tempo will change in proportion to the speed change. At 33 RPM it would already be musically flat as the correct speed is 331/3 RPM. A 1 kHz test tone - common on test records - would, at 33 RPM, give off $ frac {33}{33.33} text {kHz} $.



                  The sinewave would remain a sinewave but stretched in time and, therefore, a lower pitch.






                  share|improve this answer











                  $endgroup$




                  Say the record was mean to play sin(wt), a pure sine wave, at 33rpm, then, because of a loosen belt or any other reason, it rotates at a different RPM, how to calculate the changes in such sine wave?




                  The pitch and tempo will change in proportion to the speed change. At 33 RPM it would already be musically flat as the correct speed is 331/3 RPM. A 1 kHz test tone - common on test records - would, at 33 RPM, give off $ frac {33}{33.33} text {kHz} $.



                  The sinewave would remain a sinewave but stretched in time and, therefore, a lower pitch.







                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited 52 mins ago









                  K H

                  2,360215




                  2,360215










                  answered 1 hour ago









                  TransistorTransistor

                  88.2k785189




                  88.2k785189

























                      3












                      $begingroup$

                      To really simplify, a record has wiggles in the groove that correspond to the recorded sound pressure. (This ignores stereo, and any compounding, but it answers your question).



                      Events are recorded onto that wiggly grove as they happen -- you can think of the groove as a picture of the sound, with the time domain turned into events happening as the needle follows the groove.



                      If you play the record slower, all the events happen more slowly -- the singer sings slower and deeper, the orchestra does too, etc. Speeding it up does the opposite -- a normal recording, sped up, sounds like a hyperactive chipmunk.






                      share|improve this answer











                      $endgroup$









                      • 1




                        $begingroup$
                        I was surprised when I first learned how records work. It's so analog that it's amazing it works at all.
                        $endgroup$
                        – Toor
                        1 hour ago
















                      3












                      $begingroup$

                      To really simplify, a record has wiggles in the groove that correspond to the recorded sound pressure. (This ignores stereo, and any compounding, but it answers your question).



                      Events are recorded onto that wiggly grove as they happen -- you can think of the groove as a picture of the sound, with the time domain turned into events happening as the needle follows the groove.



                      If you play the record slower, all the events happen more slowly -- the singer sings slower and deeper, the orchestra does too, etc. Speeding it up does the opposite -- a normal recording, sped up, sounds like a hyperactive chipmunk.






                      share|improve this answer











                      $endgroup$









                      • 1




                        $begingroup$
                        I was surprised when I first learned how records work. It's so analog that it's amazing it works at all.
                        $endgroup$
                        – Toor
                        1 hour ago














                      3












                      3








                      3





                      $begingroup$

                      To really simplify, a record has wiggles in the groove that correspond to the recorded sound pressure. (This ignores stereo, and any compounding, but it answers your question).



                      Events are recorded onto that wiggly grove as they happen -- you can think of the groove as a picture of the sound, with the time domain turned into events happening as the needle follows the groove.



                      If you play the record slower, all the events happen more slowly -- the singer sings slower and deeper, the orchestra does too, etc. Speeding it up does the opposite -- a normal recording, sped up, sounds like a hyperactive chipmunk.






                      share|improve this answer











                      $endgroup$



                      To really simplify, a record has wiggles in the groove that correspond to the recorded sound pressure. (This ignores stereo, and any compounding, but it answers your question).



                      Events are recorded onto that wiggly grove as they happen -- you can think of the groove as a picture of the sound, with the time domain turned into events happening as the needle follows the groove.



                      If you play the record slower, all the events happen more slowly -- the singer sings slower and deeper, the orchestra does too, etc. Speeding it up does the opposite -- a normal recording, sped up, sounds like a hyperactive chipmunk.







                      share|improve this answer














                      share|improve this answer



                      share|improve this answer








                      edited 51 mins ago









                      K H

                      2,360215




                      2,360215










                      answered 1 hour ago









                      TimWescottTimWescott

                      6,5991416




                      6,5991416








                      • 1




                        $begingroup$
                        I was surprised when I first learned how records work. It's so analog that it's amazing it works at all.
                        $endgroup$
                        – Toor
                        1 hour ago














                      • 1




                        $begingroup$
                        I was surprised when I first learned how records work. It's so analog that it's amazing it works at all.
                        $endgroup$
                        – Toor
                        1 hour ago








                      1




                      1




                      $begingroup$
                      I was surprised when I first learned how records work. It's so analog that it's amazing it works at all.
                      $endgroup$
                      – Toor
                      1 hour ago




                      $begingroup$
                      I was surprised when I first learned how records work. It's so analog that it's amazing it works at all.
                      $endgroup$
                      – Toor
                      1 hour ago











                      2












                      $begingroup$

                      Changing the speed of the platter simply affects how fast the groove is moving under the needle, nothing else.



                      A sine wave with the time axis compressed or expanded is still a sine wave. In fact, since the groove is a direct mechanical representation of the original complex waveform, you still get the same waveform simply compressed or expanded in time.






                      share|improve this answer









                      $endgroup$













                      • $begingroup$
                        Right, agreed, in the ideal case a a sine wave of same amplitude, but with a diferent frequency, right? The point is for a sin(wt), how the change of rotation speed will affect the frequency?
                        $endgroup$
                        – Gabriel Santos
                        1 hour ago










                      • $begingroup$
                        It's linear -- double the speed means double the frequency. That's what compressing the time axis means.
                        $endgroup$
                        – Dave Tweed
                        1 hour ago
















                      2












                      $begingroup$

                      Changing the speed of the platter simply affects how fast the groove is moving under the needle, nothing else.



                      A sine wave with the time axis compressed or expanded is still a sine wave. In fact, since the groove is a direct mechanical representation of the original complex waveform, you still get the same waveform simply compressed or expanded in time.






                      share|improve this answer









                      $endgroup$













                      • $begingroup$
                        Right, agreed, in the ideal case a a sine wave of same amplitude, but with a diferent frequency, right? The point is for a sin(wt), how the change of rotation speed will affect the frequency?
                        $endgroup$
                        – Gabriel Santos
                        1 hour ago










                      • $begingroup$
                        It's linear -- double the speed means double the frequency. That's what compressing the time axis means.
                        $endgroup$
                        – Dave Tweed
                        1 hour ago














                      2












                      2








                      2





                      $begingroup$

                      Changing the speed of the platter simply affects how fast the groove is moving under the needle, nothing else.



                      A sine wave with the time axis compressed or expanded is still a sine wave. In fact, since the groove is a direct mechanical representation of the original complex waveform, you still get the same waveform simply compressed or expanded in time.






                      share|improve this answer









                      $endgroup$



                      Changing the speed of the platter simply affects how fast the groove is moving under the needle, nothing else.



                      A sine wave with the time axis compressed or expanded is still a sine wave. In fact, since the groove is a direct mechanical representation of the original complex waveform, you still get the same waveform simply compressed or expanded in time.







                      share|improve this answer












                      share|improve this answer



                      share|improve this answer










                      answered 1 hour ago









                      Dave TweedDave Tweed

                      123k9152266




                      123k9152266












                      • $begingroup$
                        Right, agreed, in the ideal case a a sine wave of same amplitude, but with a diferent frequency, right? The point is for a sin(wt), how the change of rotation speed will affect the frequency?
                        $endgroup$
                        – Gabriel Santos
                        1 hour ago










                      • $begingroup$
                        It's linear -- double the speed means double the frequency. That's what compressing the time axis means.
                        $endgroup$
                        – Dave Tweed
                        1 hour ago


















                      • $begingroup$
                        Right, agreed, in the ideal case a a sine wave of same amplitude, but with a diferent frequency, right? The point is for a sin(wt), how the change of rotation speed will affect the frequency?
                        $endgroup$
                        – Gabriel Santos
                        1 hour ago










                      • $begingroup$
                        It's linear -- double the speed means double the frequency. That's what compressing the time axis means.
                        $endgroup$
                        – Dave Tweed
                        1 hour ago
















                      $begingroup$
                      Right, agreed, in the ideal case a a sine wave of same amplitude, but with a diferent frequency, right? The point is for a sin(wt), how the change of rotation speed will affect the frequency?
                      $endgroup$
                      – Gabriel Santos
                      1 hour ago




                      $begingroup$
                      Right, agreed, in the ideal case a a sine wave of same amplitude, but with a diferent frequency, right? The point is for a sin(wt), how the change of rotation speed will affect the frequency?
                      $endgroup$
                      – Gabriel Santos
                      1 hour ago












                      $begingroup$
                      It's linear -- double the speed means double the frequency. That's what compressing the time axis means.
                      $endgroup$
                      – Dave Tweed
                      1 hour ago




                      $begingroup$
                      It's linear -- double the speed means double the frequency. That's what compressing the time axis means.
                      $endgroup$
                      – Dave Tweed
                      1 hour ago











                      0












                      $begingroup$

                      Grooves are cut with frequency correction according to RIAA equalization. Playing the record off with wrong speed increases all frequencies by the same factor (corresponding to a shift left/right on the frequency axis of the doubly logarithmic transfer function diagram). Since the frequency correction is not a straight line, this does not just result in a frequency shift but also in an uneven frequency response due to recording and replaying correction no longer being proper inverses.



                      In addition, the equalization is done in order to reduce excessive signal amplitudes on stylus and pickup. Counteracting this by wrong speed may lead to either excessive amplitudes (electrical or mechanical) or too low signals overlaid with a relatively higher noise floor.





                      share









                      $endgroup$


















                        0












                        $begingroup$

                        Grooves are cut with frequency correction according to RIAA equalization. Playing the record off with wrong speed increases all frequencies by the same factor (corresponding to a shift left/right on the frequency axis of the doubly logarithmic transfer function diagram). Since the frequency correction is not a straight line, this does not just result in a frequency shift but also in an uneven frequency response due to recording and replaying correction no longer being proper inverses.



                        In addition, the equalization is done in order to reduce excessive signal amplitudes on stylus and pickup. Counteracting this by wrong speed may lead to either excessive amplitudes (electrical or mechanical) or too low signals overlaid with a relatively higher noise floor.





                        share









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          Grooves are cut with frequency correction according to RIAA equalization. Playing the record off with wrong speed increases all frequencies by the same factor (corresponding to a shift left/right on the frequency axis of the doubly logarithmic transfer function diagram). Since the frequency correction is not a straight line, this does not just result in a frequency shift but also in an uneven frequency response due to recording and replaying correction no longer being proper inverses.



                          In addition, the equalization is done in order to reduce excessive signal amplitudes on stylus and pickup. Counteracting this by wrong speed may lead to either excessive amplitudes (electrical or mechanical) or too low signals overlaid with a relatively higher noise floor.





                          share









                          $endgroup$



                          Grooves are cut with frequency correction according to RIAA equalization. Playing the record off with wrong speed increases all frequencies by the same factor (corresponding to a shift left/right on the frequency axis of the doubly logarithmic transfer function diagram). Since the frequency correction is not a straight line, this does not just result in a frequency shift but also in an uneven frequency response due to recording and replaying correction no longer being proper inverses.



                          In addition, the equalization is done in order to reduce excessive signal amplitudes on stylus and pickup. Counteracting this by wrong speed may lead to either excessive amplitudes (electrical or mechanical) or too low signals overlaid with a relatively higher noise floor.






                          share











                          share


                          share










                          answered 1 min ago







                          user217611





































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