Shorter final drive ratio

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Old Sep 12, 2020 | 01:14 PM
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Lower final drive ratio

Based on the specs released by Acura, it looks like the gear ratios for the 2G TLX is identical to the 3G RDX, but with a lower final drive. This seems to suggest the 2G transmission will be the same, just with a different diff.

A lower final drive ratio means a higher top speed but less torque to the wheels, right? I wonder what that'll mean for the acceleration numbers... I guess at least this means freeway MPGs should be better since the engine will be spinning at a lower RPM for the same speed (edit: just noticed the EPA rating is out, and it's 22 city / 31 hwy, so same hwy as the outgoing V6 and 2 better in the city, and 1 less in both than the lighter Accord 2.0T which makes sense).

RDX: https://assets.acuratrainingguide.co...TURES_v2_0.pdf
TLX: http://www.urvi.net/forumfiles/SB/20...20Features.PDF

Edit 2: Dang I meant lower final drive in the title, not shorter. Shorter would be a higher final drive. If a mod could help update that to avoid confusion, that would be swell

Last edited by fiatlux; Sep 12, 2020 at 01:22 PM.
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Old Sep 12, 2020 | 01:54 PM
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They might have engineered the difference in the final drive ratio to compensate for something else (weight, aerodynamics, rolling resistance, etc) but the perceptible difference to the driver will be negligible.

What will be interesting is how they’re going to “beef up” the transmission for the Type S.
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Old Sep 12, 2020 | 05:07 PM
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The RDX and TLX roll on different diameter tire/wheel (just used the standard tire/wheel, assuming someone at Honda matched up the

Tire/wheel
RDX 19" + (235mm*0.55*2/25.4) = 29.17" diameter
TLX 18" + (235mm*0.50*2/25.4) = 27.25" diameter

Final drive to force calculation
TLX 3.69 / (27.25/(2*12)) = 3.25 pounds of force at tire contact patch per ftlb of final drive torque
RDX 4.17 / (29.17/(2*12)) = 3.43 pounds "

TLX/RDX 3.25/3.43 = 0.948


So if my math is right the TLX will have a ~5.2% lower force at the wheel from final drive/tire diameter overall compared to a RDX with the same torque. Since they have the same engine/transmission then that is a overall diff in torque multiplier

Last edited by Legend2TL; Sep 12, 2020 at 05:09 PM.
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Old Sep 12, 2020 | 05:18 PM
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That looks right, although final drive for the TLX is 3.59, not 3.69, so it ends up being about 7.9% lower in wheel torque.
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Old Sep 12, 2020 | 05:49 PM
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Originally Posted by Legend2TL
The RDX and TLX roll on different diameter tire/wheel (just used the standard tire/wheel, assuming someone at Honda matched up the

Tire/wheel
RDX 19" + (235mm*0.55*2/25.4) = 29.17" diameter
TLX 18" + (235mm*0.50*2/25.4) = 27.25" diameter

Final drive to force calculation
TLX 3.59 / (27.25/(2*12)) = 3.16 pounds of force at tire contact patch per ftlb of final drive torque
RDX 4.17 / (29.17/(2*12)) = 3.43 pounds "

TLX/RDX 3.16/3.43 = 0.921


So if my math is right the TLX will have a ~8.2% lower force at the wheel from final drive/tire diameter overall compared to a RDX with the same torque. Since they have the same engine/transmission then that is a overall diff in torque multiplier
corrected
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Old Sep 12, 2020 | 06:10 PM
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Originally Posted by Legend2TL

o if my math is right the TLX will have a ~7.9% lower force at the wheel from final drive/tire diameter overall compared to a RDX with the same torque. Since they have the same engine/transmission then that is a overall diff in torque multiplier

corrected
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Old Sep 12, 2020 | 09:58 PM
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Originally Posted by Legend2TL
o if my math is right the TLX will have a ~8.2% lower force at the wheel from final drive/tire diameter overall compared to a RDX with the same torque. Since they have the same engine/transmission then that is an overall diff in torque multiplier

corrected
corrected by the grammar
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