Drivetrain 5th Gear Ratio

PonyGTrider

5 Year Member
Feb 27, 2019
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Mexico
Hi all
I have a question what 5th gear ratio do I have if I replaced my T5 original gear set with the Z specs gearing set with the exception of fifth gear?

1st. 2nd. 3rd. 4th. 5th
T5 WC 3.35 1.99 1.33 1.00 0.68
T5 Z Spec 2.95 1.94 1.34 1.00 ?

I asked somewhere else and they told me with the Z specs gears set and my stock 5th gear it renders a .59 5th gear ratio is that true?
I know that my stock T5 WC came with .68 and the Z spec T5 came with .63

How can I verify if I have a .59 ratio 5th gear???

Thank you all
 
Ok I just lifted the rear and did 1/2 turn on the crank. Tha shaft did not complete a full turn. The pic below shows the indicador where the turn started an a silver mark before before completing that turn and I know I did half of the turn to the crank because my damper has laser marks at TDC (0 or 360) - 270 - 180 - and 90 and I went from TDC to 180.
So what is my 5th gear ratio? Will it be ok to asume my crank is making 0.59 turns for 1 turn of the shaft? Is that something close to 0.59:1 ratio?
Comments/suggestion will be appreciated
 

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You need to calculate or measure the circumference of the drive shaft. Put a pointer on it somehow (wire coat hanger comes to mine secured between the trans and the exhaust hanger plate), put a mark on the DS where the pointer is, and then put a mark on where the pointer is when you turn the motor over one revolution. Measure how far the DS turns (mark to mark in the direction of rotation).

Distance the DS moved / DS circumference = gear ratio of the gear the transmission was in when engine was rotated
 
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Why does the circumference matter?( Honest question)

It shouldn't matter if it's 2" or 8".... The yoke is still the yoke and 1 revolution is 1 revolution...but then again I was never good at the maths growing up ..

Wouldnt it be easier just to calculate it based of RPM, tire diameter and rear end ratio?

Effective Ratio=(MPH x rear Gear Ratio x 336) ÷ (Tire Diameter in. x RPM)
 
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Why does the circumference matter?( Honest question)

It shouldn't matter if it's 2" or 8".... The yoke is still the yoke and 1 revolution is 1 revolution...but then again I was never good at the maths growing up ..

Wouldnt it be easier just to calculate it based of RPM, tire diameter and rear end ratio?

Effective Ratio=(MPH x rear Gear Ratio x 336) ÷ (Tire Diameter in. x RPM)
I was wondering that too…
 
So with a transmission each revolution of the motor is either reduced or increased at the tail shaft via whatever gear the transmission is in. So if you turn the motor one revolution and the DS turns one time then you are 1:1 i.e. 3d gear in an AOD or 4th gear with a T5. If the motor turns one time, the transmission is in 5th gear, and the DS turns 1.47 times then its 0.68:1 fifth.

I would measure the circumference of the DS so that I can figure how much the DS turned in decimals. This is just how my brain works and how it makes sense to me.

So rotate the motor so that it makes one full turn, figure the number of turns the driveshaft made, and then divide the one full turn of the motor by the number of turns the DS made. Example would be as follows:

1/1.587 = 0.63 ratio
1/1.47 = 0.68 ratio
1/1.25 = 0.80 ratio
 
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You need to calculate or measure the circumference of the drive shaft. Put a pointer on it somehow (wire coat hanger comes to mine secured between the trans and the exhaust hanger plate), put a mark on the DS where the pointer is, and then put a mark on where the pointer is when you turn the motor over one revolution. Measure how far the DS turns (mark to mark in the direction of rotation).

Distance the DS moved / DS circumference = gear ratio of the gear the transmission was in when engine was rotated
Ok I did your method and trying to make some sense, here are the numbers:
DS circumference: 9.5”
Pointer distance traveled with one full crankshaft revolution: 16.25
So 16.25 / 9.5 = 1.711…. And then what’s the 5th gear ratio? I’m confused
 
So with a transmission each revolution of the motor is either reduced or increased at the tail shaft via whatever gear the transmission is in. So if you turn the motor one revolution and the DS turns one time then you are 1:1 i.e. 3d gear in an AOD or 4th gear with a T5. If the motor turns one time, the transmission is in 5th gear, and the DS turns 0.68 times then its 0.68:1 fifth.

I would measure the circumference of the DS so that I can figure how much the DS turned in decimals. This is just how my brain works and how it makes sense to me.

Another way would be to rotate the motor so that the DS make one full turn figure the number of turns the motor made and then divide the one full turn of the DS by the number of turns the motor made. Example would be as follows:

1/1.587 = 0.63 ratio
1/1.47 = 0.68 ratio
1/1.25 = 0.80 ratio
Ok then now it is making sense, with one crank revolution my 9.5” circumference driveshaft mark traveled 16.25” so 16.25/9.5=1.711 and 1/1.711=0.584 so I have what is called .59 5th gear ratio.

Thank you much
 
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If the motor turns one time, the transmission is in 5th gear, and the DS turns 0.68 times then it’s 0.68:1 fifth.

It’s easy to get confused on this and it happens to me also but I think you have this backwards. Think like this; if overdrive is going to get you further down the road at the same engine rpm, then the driveshaft has to be spinning faster in relation to the crankshaft. So in a .68 overdrive gear, the driveshaft will spin more than 1 full turn.

Actually, if you turn the driveshaft one full turn and the crankshaft turns .68. , it’s .68 overdrive
 
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It’s easy to get confused on this and it happens to me also but I think you have this backwards. Think like this; if overdrive is going to get you further down the road at the same engine rpm, then the driveshaft has to be spinning faster in relation to the crankshaft. So in a .68 overdrive gear, the driveshaft will spin more than 1 full turn.

Actually, if you turn the driveshaft one full turn and the crankshaft turns .68. , it’s .68 overdrive
That’s correct that’s what I stated after my first measurement. I did 1/2 turn on the crank and the the DS did a bit less than one turn which hypothetically if I did 1/2 turn on the crank and the DS did a complete turn the ratio would render 0.5:1.
I believe I have this completely understood
Thank you all for your input and patience!