[Systems Engineering Teachers' Mailing List] RE: syseng Digest,
Vol 27, Issue 15
Neville YOUNG
NYO at gwsc.vic.edu.au
Fri Sep 14 15:35:15 EST 2007
I'm not sure if this has already been sent. I have either sent a previous message or I have lost it, however, here is version 2.
The formula driver/driven is correct.
The error is in your interpretation of how the gear ratio works.
If you look at the example given for the 60 tooth gear driving a 30 tooth gear the gear ratio is simply 60/30 which gives a 2:1 ratio or just 2.
This means that for every revolution of the 60 tooth gear then the 30 tooth gear will turn twice. ie. if the 60 tooth gear is turning at 40 rpm then 2 x 40 rpm gives the speed of the driven gear - 80 rpm.
If the driver gear had 10 teeth and the driven gear had 50 teeth, then using Driver/Driven = 10/50 = 1:5 or just 0.2. If the 10 tooth (driver) gear were turning at 200 rpm then to find the speed of the 50 tooth gear you get 200 x 0.2 = 40 rpm.
I have never heard this question raised at any Systems sessions at the TEAV.\
Regards
Neville Young
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Sent: Fri 14/09/2007 12:00 PM
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Subject: syseng Digest, Vol 27, Issue 15
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Today's Topics:
1. Fw: Gear ratio formula (Adriana Agosta)
----------------------------------------------------------------------
Message: 1
Date: Fri, 14 Sep 2007 10:56:30 +1000
From: "Adriana Agosta" <eo at datta.vic.edu.au>
Subject: [Systems Engineering Teachers' Mailing List] Fw: Gear ratio
formula
To: "Systems and Engineering Teachers' Mailing List"
<syseng at edulists.com.au>
Message-ID: <001801c7f66a$39597a00$8301a8c0 at teav.local>
Content-Type: text/plain; charset="iso-8859-1"
----- Original Message -----
From: Judy Moore
To: Adriana Agosta
Sent: Friday, September 14, 2007 9:50 AM
Subject: Fw: Gear ratio formula
----- Original Message -----
From: Barton, Paul S
To: admin at datta.vic.edu.au
Sent: Wednesday, September 12, 2007 8:29 PM
Subject: Gear ratio formula
I have been working with Lorraine Tran and Kristin Allen from the VCAA in trying to resolve the issues with the gear ratio formula that was used in the 2006 exam in question 4h and 4i. In the 2006 assessment report the formula used was driver over driven for both questions. This resulted in the wrong gear ratio for 4h answer given 5:1, should be 1:5. For every one turn of the driver (60 tooth) : 5 turns of the driven (12 tooth), or large gear drives a small gear faster. The correct formula is driven over driver. This is obvious in the answer given in 4i. The incorrect answer was that a 4,800 toothed gear be fitted to a 1200RPM electric motor as the driver and then meshed with a 12 toothed driven gear to produce a 3 RPM output speed. Ignoring the practicalities of trying to create this scenario in reality, this is embarrassing that due to this incorrect formula being used the larger gear is calculated to be placed onto the shaft of the electric motor. To stay with the scenari!
o the small gear has to be fitted to the electric motor and the larger gear to the output shaft.
Kristin Allen has pointed out to me that she must follow the 2007 Study Design in which on page 37 it states:
Gear ratio A mathematical representation of the ratio of gear teeth in gear trains (e.g. 60
teeth gear drives 30 teeth gear, 60:30 or 2:1). (driver over driven).
This is incorrect, 2 turns of the 60 gear does not result in 1 of the driven and if the output speed= input speed/ VR, the wrong speed would be calculated. The correct formula is DRIVEN / DRIVER. Lorraine Tran asked why no one has raised this before but I've heard lots of people raise this at the TEAV systems study design units 1&3.
Please return your opinions quickly as our students will be sitting this exam very soon.
Thank you,
Paul Barton, Systems Engineering teacher, Dromana Secondary College.
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