[Systems Engineering Teachers' Mailing List] RE: syseng Digest,
Vol 27, Issue 17
Roger Knight
RKnight at stpats.vic.edu.au
Sat Sep 22 06:07:18 EST 2007
Very true Neville. However, what happens when we apply Driver / Driven as a method for calculating gear ratios in the following scenario? (bold italics mine) Web reference :
http://www.chevyhiperformance.com/techarticles/148_0208_gear_ratio_calculating/
An automobile uses gear ratios in both the transmission and the drive axle to multiply power. The two ratios multiplied together equal the final drive ratio. Spend a few minutes in any bench-racing session and soon you'll hear rear axle gear ratios discussed. For many performance cars, 3.73s and 4.10s are common gear choices. The rearend gear ratio refers to the relationship between the ring gear and the pinion gear. By simply dividing the ring gear (driven) tooth count by the pinion gear (driving) tooth count, the ratio is determined. For example, if we divide a ring gear with 41 teeth by a pinion gear with 10 teeth we find that the gear ratio is 4.10:1 Driven / Driver = (41/10 = 4.10).
Because transmissions are comprised of several gear choices, the transmission allows the vehicle to accelerate quickly with lower gears and to maintain a cruising rpm using higher gears. In the '60s and '70s, most transmissions offered three or four gears with a 1:1 high gear. Using a TH400 as an example, First gear is 2.48:1, Second gear is 1.48:1, and Third gear is 1:1. Multiplying the 2.48 First gear by the 4.10 rear axle results in a final drive ratio of 10.16:1 (2.48 x 4.10 = 10.16). For most street performance applications, a 10:1 final First gear ratio is usually considered optimal. The disadvantage of operating a 4.10:1 axle ratio on the street with a 1:1 high gear is excessive freeway engine speed.
Earlier, I stated that the reciprocal was applied for reasons of expression. I think there might be more to it than that. As seen here when more than one gear ratio is used to achieve a final drive in a gear train the results of NOT applying Driven / Driver would be disastrous, you'll agree!?
Roger Knight
Technology Dept.
St Patrick's College
1431 Sturt Street
BALLARAT 3350 VIC
5331 1688
rknight at stpats.vic.edu.au
________________________________
From: syseng-bounces at edulists.com.au on behalf of Neville YOUNG
Sent: Mon 9/17/2007 9:31 AM
To: syseng at edulists.com.au
Subject: RE: syseng Digest, Vol 27, Issue 17
I stick by the gear ratio for a 60 tooth driver gear and a 30 tooth driven gear as being a ratio of 2:1. There are twice as many teeth on the Driver gear compared to the Driver gear. It is a ratio describing the relationship between the number of gear teeth on the driver and driven gears and a ratio of 2:1 means that there are twice as many teeth on the driver gear. If the driven gear has 45 teeth the driver will have 90 teeth. This is the gear ratio refers to teeth on the gears and is not the speed ratio, it as David points out is 1:2.
Neville Young
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From: syseng-bounces at edulists.com.au on behalf of syseng-request at edulists.com.au
Sent: Sat 15/09/2007 12:00 PM
To: syseng at edulists.com.au
Subject: syseng Digest, Vol 27, Issue 17
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Today's Topics:
1. RE: syseng Digest, Vol 27, Issue 15 (Roger Knight)
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Message: 1
Date: Sat, 15 Sep 2007 01:22:27 +1000
From: "Roger Knight" <RKnight at stpats.vic.edu.au>
Subject: [Systems Engineering Teachers' Mailing List] RE: syseng
Digest, Vol 27, Issue 15
To: "Systems and Engineering Teachers' Mailing List"
<syseng at edulists.com.au>
Message-ID:
<EC8B0ED0F258A94498B697367C2913A2DE6E66 at Shalimar.stpats.vic.edu.au>
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Actually gentlemen both are correct ... depending on how you want to expess the gear ratio in terms of the relationship of one gear to the other.
Neville, you are correct in saying that the 60 tooth gear driving the 30 tooth gear gives us 60/30 or Driver/Driven. You then went on to say that this was a gear ratio of 2:1. Correct again ... but unconventional. The correct, or conventional expression of this gear ratio is 1:2 meaning 1 turn of the 60 tooth gear gets you 2 turns of the 30 tooth gear, providing the 60 tooth gear is the driver. Confusion creeps in here when we take 2/1 as 2:1. It is at this point when the reciprocal rule is applied as referred to by David Fletcher. Actually, I would prefer to call it an inversion. 60/30 broken down to 2/1 is actually referred to by engineers as a gear ratio of 1:2 because we conventionally refer to what the driving gear is doing first.
Next Neville referred 60/30 as 2. This is also correct if you want to calculate the output speed using the formula :
Output speed = Input speed x VR
... which only works if you used Driver/Driven to calculate VR (VR being velocity ratio or gear ratio).
Convention states that we calculate output speed using the formula :
Output speed = Input speed / VR
... which only works if you used Driven/Driver to calculate VR. It just depends on how you like it said. I think Driver/Driven is more mathematically sound because it is the root formula. But I like Driven/Driver because of the way it sounds. To invert or not to invert?? This is the question.
________________________________
From: syseng-bounces at edulists.com.au on behalf of Neville YOUNG
Sent: Fri 9/14/2007 3:35 PM
To: syseng at edulists.com.au
Subject: RE: syseng Digest, Vol 27, Issue 15
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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From: syseng-bounces at edulists.com.au on behalf of syseng-request at edulists.com.au
Sent: Fri 14/09/2007 12:00 PM
To: syseng at edulists.com.au
Subject: syseng Digest, Vol 27, Issue 15
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Today's Topics:
1. Fw: Gear ratio formula (Adriana Agosta)
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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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