Transmissons - Page 2

JJ68 · 11 Jan 2011, 20:33 (Ref:2814394)

ok so i have now managed to calulate reaching 220mph in 6th gear using the formula with trial an error, below is my working out, correct to 3 d.p.

Overall gear ratio: 1.2 x 1.07 x 2.057 = 2.641

Output speed: 1200/2.641 = 4543.733 RPM
Output speed: 4543.733 RPM /60 = 75.728 Revs/sec

16 x 25.4 = 406.4
406.4/1000 = 0.4064
Pi x 0.4064 = 1.3

Speed: 75.728 x 1.3 = 98.446 m/s

98.446 x 3600/1000 = 354.405 KM/H

354.405/1.609 = 220.264 MPH
It looks ok to me, any comments?

phoenix · 11 Jan 2011, 20:53 (Ref:2814405)

Not sure where you are getting some of your figures - you are getting there though, so stick with it.

Quote:

Originally Posted by JJ68

ok so i have now managed to calulate reaching 220mph in 6th gear using the formula with trial an error, below is my working out, correct to 3 d.p.

Overall gear ratio: 1.2 x 1.07 x 2.057 = 2.641
Great - but where is the 1.6 ratio given in the problem? And what has happened to the 14/44 final drive ratio?
Output speed: 1200/2.641 = 4543.733 RPM
Typo - should be 12000/2.641......
Output speed: 4543.733 RPM /60 = 75.728 Revs/sec
As you need to work out MPH, why convert to revs per second?
16 x 25.4 = 406.4
Where does the '16' come from?
406.4/1000 = 0.4064
Pi x 0.4064 = 1.3
If the '16' is wrong so are the above calculations.... The circumference of a 26" tyre is 2.075 metres. Effective rolling circumference will be different, as the tyre will deflect under load

Speed: 75.728 x 1.3 = 98.446 m/s

98.446 x 3600/1000 = 354.405 KM/H

354.405/1.609 = 220.264 MPH

Now really I am lost...

It looks ok to me, any comments?

JJ68 · 12 Jan 2011, 12:37 (Ref:2814654)

In reply to Phoenix,

the figure of 1.6 for the primary ratio was used in an example calculation, nowhere in the brief does it mention what ratio can be used, likewise with the diff ratio. The Diff ratio i used was, 28.8/14

Yes 1200 should read 12000.

We converted the example equation to rev/sec therfore i thought this had to be done - how would i go about directly calculating the MPH then?

Lastly the tyre size i was told to use was 16-26/15..... does this mean the calc should of been:

26 x 25.4 = 660.4
660.4 / 1000 = 0.6604
Pi x 0.6604 = 2.075

As you say the tyre circumferance completely throws my equation out of the window!

Comments?

phoenix · 12 Jan 2011, 13:14 (Ref:2814671)

Quote:

Originally Posted by JJ68

the tyre size i was told to use was 16-26/15..... does this mean the calc should of been:

26 x 25.4 = 660.4
660.4 / 1000 = 0.6604
Pi x 0.6604 = 2.075

As you say the tyre circumferance completely throws my equation out of the window!

Comments?

Yes!

or you could try

26 x pi / 36 /1760 x 1000 = 1.28933 miles per thousand wheel rpm (which is 775.597 revs per mile)

(36 inches to the yard, 1760 yards to the mile)

Or, you could use a tyre manufacturers spec sheet, like the Avon one I suggested, which states that their 26 inch tyres rotate 782 times per mile. This is because the rolling radius is always less than the radius of the tyre off the car - due to the flat bit at the bottom!

phoenix · 12 Jan 2011, 13:20 (Ref:2814673)

Quote:

Originally Posted by JJ68

In reply to Phoenix,

the figure of 1.6 for the primary ratio was used in an example calculation, nowhere in the brief does it mention what ratio can be used, likewise with the diff ratio. The Diff ratio i used was, 28.8/14

Can you cut a gear with 28.8 teeth on it? I don't think so!

In the real world you must use whole numbers of teeth, even if you don't get the 'ideal' ratio.

There is also a practical minimum to the minimum number of teeth. 14 is ok, but you couldn't use 3, for example. Smallest number of final drive pinion teeth I have come across is 7 and the highest is 17 - but these are mostly in production gearboxes/transaxles as I have only ever owned one Hewland which had 7 pinion teeth.

GNWT · 13 Jan 2011, 12:49 (Ref:2815227)

http://www.tz350.net/speedcharts_and...tm#gary_thomas
Have a look at this and see if it helps?
Any questions please ask.
Gary

JJ68 · 14 Jan 2011, 00:15 (Ref:2815479)

I appreciate the help Gary, the links on the TZ250 website look very helpful however from the three available links i cant get one of them to work..

GNWT · 14 Jan 2011, 14:15 (Ref:2815681)

Oh dear, I have tried too and no success!
Let me have your email address and I will send you a copy.

red eye · 14 Jan 2011, 23:07 (Ref:2815890)

Quote:

Originally Posted by JJ68

In reply to Phoenix,

the figure of 1.6 for the primary ratio was used in an example calculation, nowhere in the brief does it mention what ratio can be used, likewise with the diff ratio. The Diff ratio i used was, 28.8/14

Yes 1200 should read 12000.

We converted the example equation to rev/sec therfore i thought this had to be done - how would i go about directly calculating the MPH then?

Lastly the tyre size i was told to use was 16-26/15..... does this mean the calc should of been:

26 x 25.4 = 660.4
660.4 / 1000 = 0.6604
Pi x 0.6604 = 2.075

As you say the tyre circumferance completely throws my equation out of the window!

Comments?

primary ratio is for bikes so not needed for this brief.
I calculate 219.65mph using 3.14285714 fd and 1.344827586 gear ratio.

phoenix
Something I just thought of, say 1st was 7/10 and 6th was 10/20. the input shaft and lay shaft would not be square. ie the shafts at 1st gear end would be close together and at 6th gear end they would be far apart. I believe there are certain ratios that can be used together if so how do you work this out? or am I talking crap?

red eye · 14 Jan 2011, 23:09 (Ref:2815891)

Quote:

Originally Posted by GNWT

http://www.tz350.net/speedcharts_and...tm#gary_thomas
Have a look at this and see if it helps?
Any questions please ask.
Gary

spreadsheet not found

phoenix · 15 Jan 2011, 15:54 (Ref:2816094)

Quote:

Originally Posted by red eye

phoenix
Something I just thought of, say 1st was 7/10 and 6th was 10/20. the input shaft and lay shaft would not be square. ie the shafts at 1st gear end would be close together and at 6th gear end they would be far apart. I believe there are certain ratios that can be used together if so how do you work this out? or am I talking crap?

You are indeed right in your thinking.

If you have two pairs of gears running on the same two parallel shafts, the sum of the radii of each pair of gears must be the same, so that their centres are equal distances apart. This means only certain combinations of gears are practically possible, but I think the solution to this problem is beyond the scope of what you have been studying and therefore beyond the scope of the question posed in your coursework.

The solution in the case you mentioned above would probably be to use a 14/20 first gear such that the sum of the radii were the same as the sixth gear pair and use a different tooth pitch on the two different gear pairs.

As an example, if you follow the link to the Hewland gear selector chart posted earlier in this thread, you will notice there are gear pairs of 17/25 17/28 17/29 17/30 17/33 17/34 17/35 and 17/36.

Logic would suggest that the 17/36 pair would not have the same centres as the 17/25 pair - but they do. However the gears have to be kept together and fitted as pairs, because the teeth of the pairs are cut to match each other. i.e. the 17 tooth gear is not interchangeable between the other gear pairs - there are 8 unique 17 tooth gears, each with different tooth patterns to match their partner gear.

red eye · 15 Jan 2011, 18:13 (Ref:2816124)

so all of hewlands gear sets for a box have the same centres but different pitch gears. But wouldt the tooth pitch alter the tooth count and not the radii (leverage) as the centres will still be the same. Its hard to imagine with out the gears in front of me.

I know your right....but why are you right? lol

Any links with more on this?

also if its not too much trouble could you check over my calcs? Im making my own spreadsheets adding to it as I progress though my course, I know how much time goes into making them. it took me all night

26 dia tyre rolling radius = 2.074m
f/d ratio= 3.14
gear ratio= 1.32
overall ratio 4.22
=219.65mph @ 12000rpm with a rev drop of 11687rpm

red eye · 15 Jan 2011, 18:51 (Ref:2816138)

scrap the above hewland says f/d= 8/31, gear=30/24 = 300mph
my spreadsheet says 191

phoenix · 15 Jan 2011, 19:29 (Ref:2816150)

Quote:

Originally Posted by red eye

scrap the above hewland says f/d= 8/31, gear=30/24 = 300mph
my spreadsheet says 191

The reason for the difference is this:

Hewland always state input gear/output gear.

So, an 8/31 final drive is a ratio of 31 divided by 8 = 3.875:1

so a 30/24 gear pair is in fact a ratio of 24 divided by 30 = 0.8:1

phoenix · 15 Jan 2011, 20:00 (Ref:2816162)

Quote:

Originally Posted by red eye

so all of hewlands gear sets for a box have the same centres but different pitch gears. But wouldt the tooth pitch alter the tooth count and not the radii (leverage) as the centres will still be the same. Its hard to imagine with out the gears in front of me.

I know your right....but why are you right? lol

I will try to explain by example.

You might guess I have worked out a solution to the coursework problem. In my solution, 3rd gear consists of a pair of gears with 23/27 teeth giving a ratio of 1.1739:1

Let's say that the distance between the input shaft and output shaft on the gearbox is 70mm, so on a 1:1 ratio you would expect each gear to have a nominal radius of 35mm. For the ratio I have selected the nominal radii will be 32.2mm and 37.8mm which added together equals 70mm. You can check for yourself that the ratio of this pair of gears is exactly 1.1739:1 by dividing the radii.

For sixth gear I have selected 29/22 - and overdrive of .7586:1. The nominal radius of these two gears are 39.8mm and 30.2mm which added together also equal 70mm. Again you can check this ratio by dividing the radii - 30.2 by 39.8.

To get these gears to mesh, the third gear pair need a tooth pitch of 4.398mm and the sixth gear pair a tooth pitch of 4.312mm.

The tooth pitch is found by dividing the circumference of each gear by the number of teeth required.

red eye · 16 Jan 2011, 15:44 (Ref:2816394)

tooth pitch is found by dividing the circumference of each gear by the number of teeth required.

so the distace is 70mm half of that is 35mm

if i wanted a ratio of 20/30 and to make thing easy both gears 35 dia (35*pi) 109.95 circumference

35/20= 1.75 tooth pitch
35/30= 1.16

or is it 35+20=55. 109.95/55= 1.999 pitch

sorry I am crap at maths!

phoenix · 16 Jan 2011, 15:56 (Ref:2816397)

No - you haven't got it yet! If you had two gears with different two pitches they wouldn't mesh. The tooth pitches have to be the same on both gears.

For a 20/30 ratio the gear radii would be 28mm and 42mm with circumferences of 87.96mm and 131.95mm respectively.

20/30 is a ratio of 1.5:1. 28mm/42mm is also a ratio of 1.5:1. Got it?

tooth pitches would be 47.96/20 and 131.95/30. Both work out as 4.398mm.

JJ68 · 17 Jan 2011, 17:28 (Ref:2816899)

Can anyone confirm if a car gearbox has a primary ratio or not?

In the example we did in my lecture it was based on working out the gear ratios for a racing car. Which had a primary ratio of 1.6. i have since been told that a car gearboxes dont have a primary ratio! No is my lecturer trying to confuse us my including a primary ratio?

So to find 220mph i did the following

24/16 x 34/12 = 4.25

12000/4.25 = 2823.529412

2823.529412/60 = 47.0588

47.0588 x 2.074 = 97.6

97.6 x (3600/1000) = 351.36 KM/H

351.36/1.609 = 218.37 MPH

This is the closest i can get using gearbox and f/d ratios from the hewland FTR range. I now beleive this is the correct method!

This is so frustrating going round in circles....

tristancliffe · 18 Jan 2011, 11:54 (Ref:2817270)

Car drivetrains have the main gear ratio and a differential ratio. With chain or belt drives (e.g. bike engined cars) there will also be an extra ratio of the chain sprockets (and it may well be 1.6 in the example your lecturer used).

You can add that the tyre circumference will change - partly due to the weight of the vehicle [and downforce?] squashing the tyre down (reducing circumference), but also the expansion of the tyre due to the high rotation speeds (increasing diameter), so your value of 218.37mph at 12000rpm could be anywhere between 215 and 223mph depending on the car and the tyre construction.

red eye · 20 Jan 2011, 15:07 (Ref:2818339)

Quote:

Originally Posted by phoenix

No - you haven't got it yet! If you had two gears with different two pitches they wouldn't mesh. The tooth pitches have to be the same on both gears.

For a 20/30 ratio the gear radii would be 28mm and 42mm with circumferences of 87.96mm and 131.95mm respectively.

20/30 is a ratio of 1.5:1. 28mm/42mm is also a ratio of 1.5:1. Got it?

tooth pitches would be 47.96/20 and 131.95/30. Both work out as 4.398mm.

Thank you, I got it now dia*pi/teeth=pitch

just got to work out how to find the gear dia for a gear ratio.

tristancliffe · 20 Jan 2011, 16:50 (Ref:2818388)

Just invert the equation?

dia=pitch*teeth/pi

phoenix · 20 Jan 2011, 22:09 (Ref:2818545)

Quote:

Originally Posted by red eye

just got to work out how to find the gear dia for a gear ratio.

and if the above isn't the answer you were looking for, but you want to know how to calculate the diameter of each of the gears in a pair, try this:

In my previous examples I use 70mm as the distance between the input shaft and the output shaft in the gearbox, in between which the two gears must fit.

So, we know that the sum of the radii of every pair of gears must equal 70mm - are you with me so far?

How do we divide up the 70mm betweeen the two gears to find the radius of each?

You can do this either using the ratio or the number of teeth you have chosen to deliver a ratio. In this example, we will use the number of teeth.

1) we want a 6th gear ratio of 1:1 and we select 28/28

We need to split the 70mm between the shaft to allow for two gears, so the calculation is:

For gear 1: 70 / (28+28) x 28=

For gear 2: 70 / (28+28) x 28=

You can do the maths and the calcultions will give the radius of each gear.

2) we want a fifth gear ratio of 1:1.1538 and we decide on 26/30 teeth.

The radius for gear 1 is 70 / (26+30) x 26=

The radius for gear 2 is 70 / (26+30) x 30=

You can do the maths and the calcultions will give the radius of each gear.

As a check, you should find that the tooth pitch to two decimals for all the gears works out to 3.93

rabbittmaster · 21 Jan 2011, 16:51 (Ref:2818885)

do we all go to the same university by any chance?

what about calculating the max tractive force, that includes the friction coefficient of 0.8 and the vehicle weight

everything on the interweb is about trains and there are no equations to help

phoenix · 21 Jan 2011, 17:16 (Ref:2818897)

1.2 x .929 x (43/13) doesn't equal 4.27!

phoenix · 21 Jan 2011, 17:17 (Ref:2818898)

Ah! You have changed your previous post.....