Weismann FTA ?

[middy]

Weismann Transmissions

Does anyone have any pictures showing the insides of the old FTA? I’d like to see the gears. Carroll Smith mentioned in one of his books that the basic differences between the two brands [Hewland and Weismann] are that Weismann does all of his shifting on the intermediate shaft rather than on the pinion shaft, and that this considerably reduces the inertial forces involved, makes the shift notably quicker and increases the life span of the engaging dogs by a notable margin.

I am curious because if you look at the following picture of a 930 of which the gears from left to right are 4-3-2-1, the shift between 4<SUP>th</SUP>/3<SUP>rd</SUP> is on the intermediate shaft (driving shaft) and the shift between 2<SUP>nd</SUP>/1<SUP>st</SUP> is on the pinion shaft (driven shaft). The reason for that is because the 4<SUP>th</SUP>/3<SUP>rd</SUP> gear ratios are overdrive - i.e. the small gears are on the pinion shaft and are too small for engagement dog rings, hence on the intermediate shaft.

Driving : Driven
Small gear : Large gear = normal drive (e.g. 16:36 or 26:27)
Large gear : Small gear = over drive (e.g. 27:26 or 31:22)

If all of Weismann shifts are on the intermediate shaft, does this mean all the gear ratios are overdrive with a funny CWP ratio? If not, how were the dog rings fitted in? :huh:



Second question:-

What are the pros and cons of overdrive gear ratios? I thought they were for economy driving but apparently Porsche 930 and also racing 962 used overdrive as tall as 0.7 and 0.61. Please enlighten me.
Thanks in advance.

 

[Roark_John]

I've never been inside one, though I will share what I know for fact, and what intuitive conjecture will arrive at...

First, the facts:
The ring and pinion is a GM 12-bolt.
The initial gearbox design was based on a Muncie top-loader.

The intuitive conjecture:
This is not built like a traditional longitudinal transaxle. The pinion shaft is completely separate from the countershaft. My guess is that feed is taken from the input stub shaft gearset. (This gear freewheels until activated by dog or synchromesh, but constantly turns the countershaft...it can be a simple or compounding design) Since all ratios would be turning the countershaft, this take-off from the input gear (either from the mainshaft, or from the mating gear on the countershaft) will always provide current ratio selection to the CWP.

To answer your second question:

Overdrive has become a pop-tech buzzword. There are overdriven gears in all sorts of geartrains, not specifically automotive. Lathes for example have some overdriven gears, as well as underdriven gears, as well as compounding these ratios for certain applications.

Overdrive is simply about allowing the geartrain to keep the powertrain in its powerband. No hocus-pocus, just manipulation of a group of spinning levers to achieve a desirable outcome. Electric motors, internal combustion engines, steam engines, etc...all have been built to produce maximum torque within a certain rev-range. That's all that any geartrain has ever been built to access.

Weismann Transmissions

Does anyone have any pictures showing the insides of the old FTA? I’d like to see the gears. Carroll Smith mentioned in one of his books that the basic differences between the two brands [Hewland and Weismann] are that Weismann does all of his shifting on the intermediate shaft rather than on the pinion shaft, and that this considerably reduces the inertial forces involved, makes the shift notably quicker and increases the life span of the engaging dogs by a notable margin.

I am curious because if you look at the following picture of a 930 of which the gears from left to right are 4-3-2-1, the shift between 4<SUP>th</SUP>/3<SUP>rd</SUP> is on the intermediate shaft (driving shaft) and the shift between 2<SUP>nd</SUP>/1<SUP>st</SUP> is on the pinion shaft (driven shaft). The reason for that is because the 4<SUP>th</SUP>/3<SUP>rd</SUP> gear ratios are overdrive - i.e. the small gears are on the pinion shaft and are too small for engagement dog rings, hence on the intermediate shaft.

Driving : Driven
Small gear : Large gear = normal drive (e.g. 16:36 or 26:27)
Large gear : Small gear = over drive (e.g. 27:26 or 31:22)

If all of Weismann shifts are on the intermediate shaft, does this mean all the gear ratios are overdrive with a funny CWP ratio? If not, how were the dog rings fitted in? :huh:



Second question:-

What are the pros and cons of overdrive gear ratios? I thought they were for economy driving but apparently Porsche 930 and also racing 962 used overdrive as tall as 0.7 and 0.61. Please enlighten me.
Thanks in advance.

 

[middy]

Re: Weismann WFA ?

Overdrive is simply about allowing the geartrain to keep the powertrain in its powerband. No hocus-pocus, just manipulation of a group of spinning levers to achieve a desirable outcome.

Yes but normally the final drive (CWP) ratio is calculated with a fixed top gear (1:1 or slight overdrive for economical cruising) at desired top speed. Then the remaining gear ratios are figured out to keep the powertrain in its powerband.

Why would a racing car need an overdrive as tall as 0.61? Why not use a taller CWP?

p.s.
I got the model name wrong.. it's WFA - not FTA sorry.

 

[Jac Mac]

Re: Weismann WFA ?

Yes but normally the final drive (CWP) ratio is calculated with a fixed top gear (1:1 or slight overdrive for economical cruising) at desired top speed. Then the remaining gear ratios are figured out to keep the powertrain in its powerband.

Why would a racing car need an overdrive as tall as 0.61? Why not use a taller CWP?

p.s.
I got the model name wrong.. it's WFA - not FTA sorry.

Why dont you draw a schematic out for yourself, it will then become very obvious why....

 

[Roark_John]

Re: Weismann WFA ?

1:1 just happens to be a mathematically convenient ratio for calculating the final drive. There are plenty of boxes that never used a 1:1 ratio, because of a lack of compounding, or even because of compounding. Small FWD car manual transaxles come to mind. (Imp, Mini, Fiat, Saab, etc)

The benefit to using a numerically higher (ie 4.10, etc...aka LOWER geared) CWP is that you get greater torque multiplication. By using a deep overdrive, you have a top gear to grab that doesn't have your engine bouncing off the rev-limiter. Numerically lower (ie 2.08, etc...aka HIGHER geared) CWPs require really low gears (granny-gears) to be useful for anything other than top-speed (Bonneville style). Not to mention the physical size of the low gears, the mass that must be spun up, the limitation of shaft diameter...it's much easier to change the gear ratio of the CWP, because the diameter stays the same on the ring gear, with the pinion either growing or shrinking.

Yes but normally the final drive (CWP) ratio is calculated with a fixed top gear (1:1 or slight overdrive for economical cruising) at desired top speed. Then the remaining gear ratios are figured out to keep the powertrain in its powerband.

Why would a racing car need an overdrive as tall as 0.61? Why not use a taller CWP?

p.s.
I got the model name wrong.. it's WFA - not FTA sorry.

 

[middy]

Re: Weismann WFA ?

Why dont you draw a schematic out for yourself, it will then become very obvious why....

OK, will do.

 

[middy]

You mean plotting a graph of mph against rpm using this formula?

mph =
rpm x 60 x 2π x RR x (Tp/Tw) x (Tdriving/Tdriven)
36 x 1760

RR= tyre rolling radius (in)
Tp= no. of teeth - pinion
Tw= no.of teeth - crown wheel
Tdriving= no. of teeth - driving gear ratio
Tdriven= no. of teeth - driven gear ratio


Yes, you're right that if a CWP ratio is altered, the same result can be achieved if the gear ratio is changed as well. But what I still don't understand is that according to the following link, overdrive is for high speed cruising, saving fuel at the cost of less torque.
Less engine torque? How can that be?

Overdrive (mechanics) - Wikipedia, the free encyclopedia)

 

[Dave Deerson]

By the time you add in the final drive numbers( like a 4.22 or numerically higher), The "overdrive" is not really that tall of a final ratio. "Overdrive" as you know it is just when a 5th or 6th gear is overdriven on the output so the final overall drive ratio is just taller( like changing a 4.22 to a 2.73) for fuel mileage or cruising.

 

[Roark_John]

First off, don't believe everything you read on Wikipedia without sources and without verifying sources. Secondly, this article isn't bad, though a bit tech-lite, even pop-tech.

Every engine develops power within a certain rev-range. Typically smaller engines (smaller displacement, fewer cylinders, or both) develop their horsepower higher in the rev-range with comparitively little torque down below. Horsepower though, is a cheated number and term. There really is no such thing. Horsepower is work done over time. Work is Torque. That is why on every dyno-graph EVER done, the HP/TQ intersection is 5252rpm. Because that is the contrived number used to make the formula for HP work.

Gears are used for torque multiplication. Gears are nothing more than spinning levers. The teeth on gears allow them to exert this leverage on their mated gear and transmit this torque from one shaft to the next.

Let's present a hypothetical. Say you have a 1.8L 4 cylinder engine with max torque of 100 ft.lbs. Your first gear ratio is 2:1, second is 1.75:1, third is 1.5:1, fourth is 1.25:1, fifth is 1:1, sixth is 0.75:1, and seventh is 0.50:1. Torque at the output (before the final drive) in first will be 200 ft.lbs., in second it will be 175 ft.lbs., etc. In sixth gear torque at the output will be 75 ft.lbs., and in seventh gear it will be only 50 ft.lbs., but the engine will hopefully be spinning in the most efficient part of its powerband.

Now, to further our hypothetical, say we have a 3.5:1 final drive ratio. That means at first gear, at the rear wheels, our 1.8L engine's 100 ft.lbs. of torque has been manipulated to a final number of 700 ft.lbs. That's a pretty healthy output and will actually allow a 100 ft.lbs. engine to move a 3500lb. car down the road without getting in its own way. Now in sixth gear (second to last gear) on the highway, in cruising mode, torque to the rear wheels will be 262.5 ft.lbs. This isn't enough torque to move a heavy mass from a dead stop very quickly, but is adequate to keep one in motion...while in the most efficient rev-range of the engine. In seventh (top or final gear) torque to the rear wheels will only be 175 ft.lbs.! Still more than direct output, but far less than with all the additional gear multiplication provided by the previous ratios.

So as you can see by our example, it's not that the engine is producing any less torque, it's that the drive wheels are seeing less torque multiplication. That's what the Wiki-article was trying to articulate, though failed miserably. Sometimes being technical, accurate and exact is the only way to communicate a point thoroughly.

This is exactly how gearing works. It's not overly complicated. The point is ALWAYS to attempt to match gearing to your engine's output and intended usage.

Yes, you're right that if a CWP ratio is altered, the same result can be achieved if the gear ratio is changed as well. But what I still don't understand is that according to the following link, overdrive is for high speed cruising, saving fuel at the cost of less torque.
Less engine torque? How can that be?

Overdrive (mechanics) - Wikipedia, the free encyclopedia)

 

[Seymour Snerd]

First off, don't believe everything you read on Wikipedia.... Horsepower though, is a cheated number and term. There really is no such thing. Horsepower is work done over time. Work is Torque. That is why on every dyno-graph EVER done, the HP/TQ intersection is 5252rpm. Because that is the contrived number used to make the formula for HP work..

1. You also have to read wikipedia to know what it says. Middy badly misquoted the article by inserting the word "engine" in front of torque where if anything he should have said "wheel" or "axle". The original statement in the wiki article is correct.
2. There's nothing "cheated" about "horsepower". It's a physical measure just as "real" as temperature, distance, time, torque, force, momentum or weight.
3. 5252 is not a "contrived" number, it's simply a consequence of how the two units are defined in the English measurement system.
4. Work is NOT torque. Work is the amount of energy transferred by a force acting through a distance. Torque is the tendency of a force to rotate an object about an axis. The fact that their English units both use the words "foot" and "pound" does not make them the same unit.
5. I don't see anything factually wrong in the wikipedia article. It does fail to distinguish explicitly among the three or more different current meanings of the word "overdrive" and thus encourages some confusion.

 

[Roark_John]

I only disagree with the highlighted posts.

A Shetland Pony is a horse. A Clydesdale is also a horse. As is a Zebra, a Mustang and a Burro. Do they all have the same output? Of course not. Horsepower is a remnant from the days when equipment was pulled by drafthorses (or draughthorses, if you prefer) to give the purchaser an idea of how much work the new steam engine would/could do, given that a farmer/miner/etc already knew how much tonnage could be pulled given a 4,6,8 & 12 horse team.

Even between two horses of the same lineage, their output will differ. However with a steam engine, diesel engine, gas engine or turbine, ONE horsepower is still ONE horsepower.

When the term was decided on, it had to be defined, then standardized. Again, I reiterate...it is a bastardized term and measurement. It is based off NO other scientific measurement. It is derived from torque using a hokey formula that is intentionally biased to arrive at this specific number...it is not reversible like any other mathematical formula.

In the same vein, all of the portable/roller based dynamometers use a bogus formula to ESTIMATE what a unit of HP is...because the math didn't work when using the established formula for HP derivation. Mustang Dyno was the first in the field for the current crop of roller style dynos. But don't take my word for it, do the research for yourself. There are several stories from the creators of the product that describe how they arrived at their own formula to allow them to arrive at their definition of HP.

Horsepower is a non-entity and is just a term to get people salivating at the stories being told. Torque does the work, and all anyone really needs to effectively gear their vehicle is a torque curve graph. Knowing what happens when, and what you want done, is more than adequate for gearing purposes. After all, that's what F1, Indy and rally teams use.

1. You also have to read wikipedia to know what it says. Middy badly misquoted the article by inserting the word "engine" in front of torque where if anything he should have said "wheel" or "axle". The original statement in the wiki article is correct.
2. There's nothing "cheated" about "horsepower". It's a physical measure just as "real" as temperature, distance, time, torque, force, momentum or weight.
3. 5252 is not a "contrived" number, it's simply a consequence of how the two units are defined in the English measurement system.
4. Work is NOT torque. Work is the amount of energy transferred by a force acting through a distance. Torque is the tendency of a force to rotate an object about an axis. The fact that their English units both use the words "foot" and "pound" does not make them the same unit.
5. I don't see anything factually wrong in the wikipedia article. It does fail to distinguish explicitly among the three or more different current meanings of the word "overdrive" and thus encourages some confusion.

 

[Seymour Snerd]

I only disagree with the highlighted posts.

A Shetland Pony is a horse. A Clydesdale is also a horse. As is a Zebra, a Mustang and a Burro. Do they all have the same output? Of course not. Horsepower is a remnant from the days when equipment was pulled by drafthorses .....

All that just because of the name of the unit?

You do realize, I hope, that not eveyone's foot is 12 inches long. Also, not all forearms have a length of one cubit.

How do you live with this?

 

[EGLITOM]

It may has been in the beginning that "hp" has been just a term to find a equvivalent to the power of one Horse. (Owning three horses i´m aware of the fact that each horse has different power). The definition of horsepower was different depending on which country and which application you are in, Within this applications to my knowledge "horsepower" approximate conversion factors has been defined.
"KW " is the unit to use for POWER. To arrive from torque to power, there is no "hookey" formular, as this is exactly definded as well and reversible.

<TABLE border=1 cellPadding=3><TBODY><TR><TD align=middle>[SIZE=+2]P = M * f * (Pi/30000) [/SIZE]<TR><TD align=middle>= [SIZE=+1]<INPUT onchange=Pberechnen() value=684 maxLength=4 size=4 name=PM {95919BE0-C436-4eab-8083-096E94826667}="684">[/SIZE]Nm * [SIZE=+1]<INPUT onchange=Pberechnen() value=3800 maxLength=4 size=4 name=Pf {95919BE0-C436-4eab-8083-096E94826667}="3800">[/SIZE]min<SUP>-1</SUP> * (3,1415/30000) = [SIZE=+1]<INPUT onfocus="blur();alert('Dieses Feld kann nicht geändert werden.')" value=272.1875875070197 maxLength=3 size=3 name=P {95919BE0-C436-4eab-8083-096E94826667}="272.1875875070197">[/SIZE]kW <INPUT onclick=Pberechnen() value=berechnen type=button> </TD></TR></TBODY></TABLE>
<TABLE border=1 cellPadding=3><TBODY><TR><TD align=middle>[SIZE=+2]M = P / f * (30000/Pi) [/SIZE]<TR><TD align=middle>= [SIZE=+1]<INPUT onchange=Mberechnen() value=305 maxLength=3 size=3 name=MP {95919BE0-C436-4eab-8083-096E94826667}="305">[/SIZE]kW / [SIZE=+1]<INPUT onchange=Mberechnen() value=5000 maxLength=4 size=4 name=Mf {95919BE0-C436-4eab-8083-096E94826667}="5000">[/SIZE]min<SUP>-1</SUP> * (30000/3,1415) = [SIZE=+1]<INPUT onfocus="blur();alert('Dieses Feld kann nicht geändert werden.')" value=582.507091716337 maxLength=4 size=4 name=M {95919BE0-C436-4eab-8083-096E94826667}="582.507091716337">[/SIZE]Nm </TD></TR></TBODY></TABLE>

Horsepower (HP) is the name of several units of measurement of power. The most common definitions equal between 735.5 and 750 watts.<SUP id=cite_ref-0 class=reference>[1]</SUP>
Horsepower was originally defined to compare the output of steam engines with the power of draft horses. The unit was widely adopted to measure the output of piston engines, turbines, electric motors, and other machinery. The definition of the unit varied between geographical regions. Most countries now use the SI unit watt for measurement of power. With the implementation of the EU Directive 80/181/EEC on January 1, 2010, the use of horsepower in the EU is only permitted as supplementary unit.

The definition of the horsepower also has varied between different applications:

• The mechanical horsepower also known as imperial horsepower of exactly 550 foot-pounds per second is approximately equivalent to 745.7 watts.
• The metric horsepower of 75 kgf-m per second is approximately equivalent to 735.499 watts.
• The boiler horsepower is used for rating steam boilers and is equivalent to 34.5 pounds (≈15.65 kilograms) of water evaporated per hour at 212 degrees Fahrenheit (=100 degrees Celsius, =373.15 Kelvin), or 9,809.5 watts.
• One imperial horsepower for rating electric motors is equal to 746 watts.
• Continental European electric motors used to have dual ratings, using conversion rate 0.735 kW for 1 HP
• The Pferdestärke PS (German translation of horsepower) is a name for a group of similar power measurements used in Germany around the end of the 19th century, all of about one metric horsepower in size.<SUP id=cite_ref-1 class=reference>[2]</SUP><SUP id=cite_ref-2 class=reference>[3]</SUP>
• The Royal Automobile Club (RAC) horsepower or British tax horsepower is an estimate based on several engine dimensions.

In regard to the overdrive definition. The ratio of gears is not defining if the gear is an "overdrive" gear. My definition would be the following. I would call an given ratio as overdrive ratio if on a given straight ( horizontal flat) street, a given vehicle ( car bike) would not be able to achieve the max RPM with the given engine power / torque output in the given gear.
I always gear my racebikes (by changing the final ratio) depending on the track. If the track has a bigger partition of long straights, i gear for max top speed. THat means the gearing which allow me to have to average highest speed on the straights. If the track has less straights i gear for acceleration. This said it depends also on the powerband characterics of your engine. My engine has a pretty narrow powerband, therefore a lot of shifting is involved to keep the thing going, therefore also the correct ratios relation is important. If your engine has wide powerband, the relation of ratios is not as critical and also the final gearing is not as important because your torque curve is able to cover a wider spread. Here i would gear for the final ration giving me the least shifting actions on track ( shifting takes time and slows you down).

TOM

 

[middy]

Hi Roark,

Got it. Thanks for the clarification :thumbsup: . By the way, you say that 700 ft.lbs output will move a 3500 lbs car from a dead stop - how do you calculate this?




Tom,

In regard to the overdrive definition. The ratio of gears is not defining if the gear is an "overdrive" gear. My definition would be the following. I would call an given ratio as overdrive ratio if on a given straight ( horizontal flat) street, a given vehicle ( car bike) would not be able to achieve the max RPM with the given engine power / torque output in the given gear.

Well said.
I know it’s a bit far-fetched but is there a formula to calculate the lowest output (in top gear) that would be able to achieve the max rpm with the given engine power/torque? I guess aerodynamic drag etc come into play? But generally, is it about 3 times more than the max engine torque?

 

[Roark_John]

Are you intentionally obtuse or just want to compare dick-size in the men's room?

Did you bother to read, then attempt to comprehend what I wrote? I stated what my problem with the issue is. You are obviously one of the sorts that think YOU are so clever because you accept what everyone else calls something and engage no critical thinking, because after all, it's gotta be right or no one else would call it so & so.

My life isn't dependent on compromise, conformity or accepting morons for what they are. Thankfully my heart beats independently of the herd, my breath is taken without permission from my neighbors, and my thoughts are not sanctioned by any other than myself.

I have contributed good, correct and ACCURATE tech to this forum. Until YOU have designed & built even ONE gearbox, please refrain from dropping your 2 cents on me. You trying to correct me from a position of ignorance and incorrectness does nothing for anyone.

All that just because of the name of the unit?

You do realize, I hope, that not eveyone's foot is 12 inches long. Also, not all forearms have a length of one cubit.

How do you live with this?

 

[Roark_John]

In our hypothetical, our 1.8L 4-cylinder engine produces a maximum of 100 ft.lbs. of torque. In first gear (a 2:1 ratio) our engine's torque has been multiplied by this ratio, with the resulting 200 ft.lbs. of torque being sent to the final drive (a 3.5:1 ratio) with that torque again being multiplied to a motive force of 700 ft.lbs. to the wheels.

It typically takes at least 1 ft.lb. of torque per 12 lbs. of mass to initiate movement. This is a rough estimate based on calculations of inertia, friction, potential energy, thermal conversion, etc. I didn't write the formulas, nor did I put pen to paper, I'm just quoting what a very experienced powertrain engineer has intimated to me.

It's this torque multiplication effect of gearing that allows a mated engine to move large loads great distances with reasonable efficiency.

Hi Roark,

Got it. Thanks for the clarification :thumbsup: . By the way, you say that 700 ft.lbs output will move a 3500 lbs car from a dead stop - how do you calculate this?

 

[middy]

I like the 930 and Weismann low ratio CWP set-up (i.e. 4.22 etc) so that gear wheel sizes can be "evenly spread" rather than granny geared. With this set up, the shifting can be put on the intermediate shaft, which is what Weismann did. Clever!

Strangely, not all Porsche gearboxes are like that though, as G50 and GT3 use high ratio CWP (3.44) and granny gears. Hewland too.



By the way, the numbers don't add up - 700 ft.lbs x 12 lbs = a lot higher than 3,500 lbs

 

[Mike]

I'm no expert here, and I certainly don't want any part of most of this conversation, but I know from the little experience that I have working with gear boxes that varying the gear sizes, even if the same combined final drive is acheived, will have a great effect upon what forces are put where within the gear box. I'm guessing that the variations you people are noticing within different boxes is do to a number of things, space constraints being right up there, but also this attempt to manipulate force within the gear box itself. This choice of how to manipulate force probably has a great deal to do with what R&P design is used and how much force capability is designed into that area of the box. Don't forget that many boxes use drop gears as a "pre" R&P ratio adjuster.

 

[Pat W]

Weismann Transmissions

Does anyone have any pictures showing the insides of the old FTA? I’d like to see the gears. Carroll Smith mentioned in one of his books that the basic differences between the two brands [Hewland and Weismann] are that Weismann does all of his shifting on the intermediate shaft rather than on the pinion shaft, and that this considerably reduces the inertial forces involved, makes the shift notably quicker and increases the life span of the engaging dogs by a notable margin.

I am curious because if you look at the following picture of a 930 of which the gears from left to right are 4-3-2-1, the shift between 4<SUP>th</SUP>/3<SUP>rd</SUP> is on the intermediate shaft (driving shaft) and the shift between 2<SUP>nd</SUP>/1<SUP>st</SUP> is on the pinion shaft (driven shaft). The reason for that is because the 4<SUP>th</SUP>/3<SUP>rd</SUP> gear ratios are overdrive - i.e. the small gears are on the pinion shaft and are too small for engagement dog rings, hence on the intermediate shaft.

Driving : Driven
Small gear : Large gear = normal drive (e.g. 16:36 or 26:27)
Large gear : Small gear = over drive (e.g. 27:26 or 31:22)

If all of Weismann shifts are on the intermediate shaft, does this mean all the gear ratios are overdrive with a funny CWP ratio? If not, how were the dog rings fitted in? :huh:



Second question:-

What are the pros and cons of overdrive gear ratios? I thought they were for economy driving but apparently Porsche 930 and also racing 962 used overdrive as tall as 0.7 and 0.61. Please enlighten me.
Thanks in advance.

A#1 The center distance on our transaxle is much larger than any other transaxle allowing us to place all the shifting elements on the primary shaft. That is where they belong.


A#2 Depending on the application Bonneville for instance all the ratios in the transmission were overdriven to get the desired speed 400 plus mph with a 31 inch tire a 2.29 gear ratio and a KB 1671 blown 510 cid engine. In this case there was not a taller final drive gear made (2.08 and 2.13 very rare) for a 12 bolt.

In any design there is a compromise. To get the low (granny gear) and the top speed (using an existing transaxle) you will need an overdrive. If you are designing from scratch you try not to do that.

pat W:thumbsup:

 

[Roark_John]

Did you drive the pinion off the front of the input/countershaft ala modified QC...or is that propietary info you're unwilling to share?

A#1 The center distance on our transaxle is much larger than any other transaxle allowing us to place all the shifting elements on the primary shaft. That is where they belong.


A#2 Depending on the application Bonneville for instance all the ratios in the transmission were overdriven to get the desired speed 400 plus mph with a 31 inch tire a 2.29 gear ratio and a KB 1671 blown 510 cid engine. In this case there was not a taller final drive gear made (2.08 and 2.13 very rare) for a 12 bolt.

In any design there is a compromise. To get the low (granny gear) and the top speed (using an existing transaxle) you will need an overdrive. If you are designing from scratch you try not to do that.

pat W:thumbsup: