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 K4B 050 Transmission
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aplekker

USA
494 Posts

Posted - 03/20/2009 :  09:57:20  Show Profile  Reply with Quote
Hi Alex,

You are right, the valve in question is not #29.

I am glad that you confirmed the situation around the venting valves. Did you get any hard copy drawings from the Classic center? If so, I would like a copy.

I added some pics in order to clarify the situation around the nickel size valve. Please let me know if we are talking about the same thing.

Here a pic of the complete oil distribution plate, with valves according to the service manual:





Valve #29 in detail:





Valve #36 (lower middle), valve in question upper left.


[br

Valve #35





Drawing from service manual:







1965 600 SWB #248
1968 6.3 #0347
1971 6.3 #5745 Euro
1979 6.9 #6857 Euro
1979 450SLC 5.0 EURO
1981 300SD
1989 560SEL
2003 CL600 Brabus T12 570HP

Edited by - paul-NL on 08/31/2017 16:30:05
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aplekker

USA
494 Posts

Posted - 03/20/2009 :  10:16:05  Show Profile  Reply with Quote
Hi Ron,

You are right about the seal damage. The input shaft's front bearing is in the coupler, so it is kind of loose trying to move it in there. If you do that in horizontal position, you take the chance that you will damage the lip of the seal with the hollow shaft from the coupler.

For any one that want to know how to properly remove and mount the coupler, here are some pics. Again, the transmission should sit on the floor in vertical position. I do it with the rear mounting bracket still attached and some wooden blocks. I am sure MB had some kind of expensive fixture.

Screw some bolts in the coupler to get a better grip:





Fluid coupler with hollow shaft:





This is what happens is you do it wrong:










1965 600 SWB #248
1968 6.3 #0347
1971 6.3 #5745 Euro
1979 6.9 #6857 Euro
1979 450SLC 5.0 EURO
1981 300SD
1989 560SEL
2003 CL600 Brabus T12 570HP

Edited by - paul-NL on 08/30/2017 12:02:22
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aplekker

USA
494 Posts

Posted - 03/24/2009 :  15:37:54  Show Profile  Reply with Quote
I am having a huge problem with some theory on this transmission. This transmission has three planetary gear sets, of which the first two are connected together and form a compound epicyclic gear train. After making a sketch on how this sticks together (no manual on that) I found a way to calculate the transmission ratio of thet set. Now the problem: according to the specs, the reductions are:
1st: 3.98
2nd: 2.46
3rd: 1.58
4th: 1

I calculated the reduction of the compound set at 1.56, which is obviously wrong (should be 2.46).
Here is the formula: ratio is (1 + (A1/S1+1)*S2/A2) / (1+S2/A2), of which is:
A1: first annulus gear, 76 teeth
S1: first sun gear, 50 teeth
A2: second annulus gear 76 teeth
S2: second sun gear 44 teeth

so ratio = (1 + (76/50 + 1) * 44/76) / (1 + 44/76)
= 2.46 / 1.58
= 1.56

The last set calculates fine at 1.58

I spent a few hours last night redoing the calculation, but always came back at 1.56 So finally this morning I stuck the gearset back together, attached brake band B1 and stuck the whole thing in a vise, thus simulating brake band B1 activated. After turning the input shaft 14 times, I got a 9 time revolutiuon of the intermediate shaft. This calculates to 14/9 = 1.55!

What is going on here? I somehow suspect that there is another interaction between the compound set and the rear planetary set, since the manual (page 27-0/1) states that the 2.46 reduction is "in front, center and rear planetary gear set".

I can't get it out of my mind, but have other things to do. Any one have any ideas, or does anyone know a manual where the three planetary gear set transmissions are explained? (I have the two planetary one).

1965 600 SWB #248
1968 6.3 #0347
1971 6.3 #5745 Euro
1979 6.9 #6857 Euro
1979 450SLC 5.0 EURO
1981 300SD
1989 560SEL
2003 CL600 Brabus T12 570HP
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paul-NL

Netherlands
4296 Posts

Posted - 03/24/2009 :  16:45:22  Show Profile  Reply with Quote
Hi Albert,

first I have to say I have no clou about those calculation, but I want just to help to find the mistake.

You calculate the second gear on 2,46 which is correct.
You calculate the third gear on 1.58 which is also correct.
But then you divide the second gear by the third gear and go wrong.

Why do you divide the second gear by third gear ???
See also Tabellenbuch 1969 page 269 and 271
Or Tabellenbuch 1972 page 286 and 287

an other Hint might be: Brakeband 1 is never activated "alone" but always in combination with Brakeband 2 ..... and then you should get i = 2,46 ...... In other words, the combination you shifted and tested is NEVER used. So you calculated it correct but that NEVER happens to be the fact ..... right ???

Edited by - paul-NL on 03/24/2009 18:22:39
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Ron B

Australia
11633 Posts

Posted - 03/24/2009 :  16:54:25  Show Profile  Visit Ron B's Homepage  Reply with Quote
ummmm ...I think it's best you leave the trans for a few days .
As long as they work I am sure the trans will be ok.
I have a color picture of the DB trans as fitted to the W111 if that is of any help.I haven't looked at it for a while but I think it indicates powerflow.

quote:
12-14-2004, 11:49 PM #8
Tom Hanson
MBCA Member

What the heck, try to stuff a MB 6.9 liter V8 in it. What a machine that would be..
__________________
Tom Hanson
Orange County Section

Edited by - paul-NL on 08/29/2017 20:39:18
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aplekker

USA
494 Posts

Posted - 03/24/2009 :  20:40:55  Show Profile  Reply with Quote
Hi Paul and Ron,

I have seen the coincedence on the 2.46 and 1.58 in the calculation of the first two planetary sets, but still cannot figure out the relevance.

Paul, where did you see that B1 will never be activated alone???
That cannot be true. It has to be activated with either B2 or K2, but never with K1.

Here is the scheme:

In 1st: B1 and B2 2.46 * 1.58
In 2nd: B1 and K2 2.46
In 3rd: K1 and B2 1.58
In 4rd: K1 and K2 1
In rev: B3

The clutch packs K1 and K2 kind of 'shorten' out the planetary gear sets, or make the reduction 1:1.

The problem I still have is that the first two planetaries somehow get to a reduction of 1.56, which cannot be true. But it somehow is proven by the test...

Ron, I agree, I left all the paperwork at work, was not going to look at it again for a while. I agree, the transmission works, but somehow I cannot handle not being able to explain. Do you know if there is a manual about the three planetary trannies, like there is about the two's? (actually it is also in the "passenger cars from 1959...").

If you have anything about the W111 tranny, is that one with three planetary sets? If so, I would love to see a picture.

In the end it will all work out, it has to. Maybe I made a mistake with the composition of the first two gear sets.. But, it worked that way on the bench. maybe I will try that again tonight,

Thanks for your reactions.

1965 600 SWB #248
1968 6.3 #0347
1971 6.3 #5745 Euro
1979 6.9 #6857 Euro
1979 450SLC 5.0 EURO
1981 300SD
1989 560SEL
2003 CL600 Brabus T12 570HP
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paul-NL

Netherlands
4296 Posts

Posted - 03/25/2009 :  07:54:59  Show Profile  Reply with Quote
Hi Albert:
according to your figures :
In 1st: B1 and B2 2.46 * 1.58
In 2nd: B1 and K2 2.46
In 3rd: K1 and B2 1.58
In 4rd: K1 and K2 1
In rev: B3

But according to the facts in my Tabellenbuch for the M100/K4B050:
In 1st: F and B2 3.98
In 2nd: B1 and B2 2.46
In 3rd: K1 and B2 1.58
In 4rd: K1 and K2 1
In rev: B3 and F 4,15
So I believe you got the wrong tabel picked out and not that from the M100gearbox.

So what you have tested is imho an non-excisting gearoption ....
Redo your test and activate both brakebands and you should get i = 2.46.
In the Tabel you can clearly see that "Brakeband 1" only is used for second gear in combination with B2. In all other gears B1 is not used.
As you can see Brakeband 1 is only used in combination with brakeband 2 for second gear.

Edited by - paul-NL on 03/25/2009 08:59:20
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aplekker

USA
494 Posts

Posted - 03/25/2009 :  10:06:00  Show Profile  Reply with Quote
Paul,

You might have hit the nail on the head here. I have no idea what "F" is, but it might be the one way clutch in the second gear set. That would also confirm my idea that somehow the two gearsets interact together.

I do not have that "Tabellenbuch", nor any other manual on the K4B 050 other than the assembly manual. Could you post a picture of the page that you are referring to?

THANKS VERY MUCH FOR YOUR REPLY, I really think this will be worked out now.

Albert

1965 600 SWB #248
1968 6.3 #0347
1971 6.3 #5745 Euro
1979 6.9 #6857 Euro
1979 450SLC 5.0 EURO
1981 300SD
1989 560SEL
2003 CL600 Brabus T12 570HP
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paul-NL

Netherlands
4296 Posts

Posted - 03/25/2009 :  10:55:57  Show Profile  Reply with Quote











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aplekker

USA
494 Posts

Posted - 03/25/2009 :  11:04:00  Show Profile  Reply with Quote
Thanks a lot, Paul, that will really help.

Will let you know later what I found out.

1965 600 SWB #248
1968 6.3 #0347
1971 6.3 #5745 Euro
1979 6.9 #6857 Euro
1979 450SLC 5.0 EURO
1981 300SD
1989 560SEL
2003 CL600 Brabus T12 570HP
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paul-NL

Netherlands
4296 Posts

Posted - 03/25/2009 :  11:34:19  Show Profile  Reply with Quote
Albert,

F will be the "Freilauf" as I believe. (see the picture from the 600 gearbox).
Hope you will now get it all together.




Edited by - paul-NL on 03/25/2009 12:29:33
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karra

Finland
868 Posts

Posted - 03/25/2009 :  12:35:41  Show Profile  Reply with Quote
quote:
Originally posted by aplekker

As a matter of fact, the K4B 050 transmission is, IMHO, a lot easier to do than a 722.3.


I decided to borrow this topic as it is very active . I used Search but could not find any information what kind of ATF is recommended for these old transmissions.

My 1968 Coupe has a bit worn out tranny because it squeeks when starting to drive uphill and also when hot it shifts to higher gear at approx. 45 MPH speed only (when cold then earlier). I have tried both Valvoline Max Life and Mobil Fully Synthetic AFT on the Coupe.

On my previous 1984 W126 I used succesfully Pennzoil Dexron III ATF as well as in my Tahoe also.

Should I find somewhere a Dexron II fluid to my 6.3 ? I think it needs the old type ATF containing the friction additives.

I guess a Dexron III type is suitable for my 1993 E420 ?

Kari Pykäläinen

1971 6.3 #5581
1968 280 SE Coupé
1993 E420
1995 Tahoe 350cid

Edited by - karra on 03/25/2009 12:37:35
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Chris Johnson

USA
3751 Posts

Posted - 03/25/2009 :  12:47:06  Show Profile  Reply with Quote
Dexron III is also appropriate for the the early transmissions, including the K4B 050.

Chris Johnson
If you aren't constantly impressed with your car, then it needs fixing.
100.012-12-000790
100.012-12-000867
www.300SE.org
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aplekker

USA
494 Posts

Posted - 03/25/2009 :  13:20:29  Show Profile  Reply with Quote
Yes, Paul, it is the "freilauf", or the free wheeling clutch, that only acts in one direction.

What really solved the problem was the sequence of the actuators, that I got wrong.

For any one interested, here is how it works:

Take a look at Paul's picture, and you will follow me (boy, I wished I had that yesterday, I had to draw it up myself). You will see the following details:
There are 3 planetary gear sets (Planetenradsatz), two in the front part of the gear set and one in the back. They have the following characteristics:

S1: 50 S2: 44 S3: 50 sun gear, number of teeth
P1: 14 P2: 17 P3: 27 planet gears, number of teeth (does not matter in calculations)
O1: 76 O2: 76 O3: 76 outer gear, number of teeth


- Input shaft connected to sun gear S1 and to inner clutch K1.
- Planet gear carrier P1 connected to drum B3 and outer clutch K1, intermediate shaft, outer gear O2, outer gear O3 (through intermediate shaft) and inner clutch K2.
- Outer gear O1 connected to planet carrier P2, to free wheeling unit F through hollow shaft
- Sun gear S2 connected to drum B1
- Drum B2 connected to outer clutch K2, sun gear S3, AND Free wheeling unit F (I missed this originally).
- Planet carrier P3 connected to output shaft.

What happens:
First gear:
- Only B2 is activated, holding drum B2 and thus sun gear S3, which makes the rear carrier a simple planetary gear train of which the outer gear O3 is driving and the planet gear P3 is driven (output shaft). Gear ratio 1 + 44/76 = 1.58
- The planet P2 and outer gear O1 are hold in place by F (through hollow shaft)against drum B2 (which I missed originally), so the first set is now a simple planetary gear train of which sun gear S1 is driving (input shaft) and planet carrier P1 is driven (output shaft). Gear ratio 1 + 76/50 = 2.52
- Total: 1.58 * 2.52 = 3.98

Second gear:
- B1 and B2 are activated. B2 same as in first gear, ratio 1.58
- B1 now holds sun gear S2, and both the front and central planetary gear sets form a compound epicyclic train, of which sun gear S1 is driving (input shaft), planet carrier P1 is driven (intermediate shaft) and sun gear S2 is stationary). Ratio is now a complex calculation, see my earlier thread. See http://www.most.gov.mm/techuni/media/ME02015_139_158.pdf for theory.
ratio: 1.56
Total ratio: 1.58 * 1.56 = 2.46


Third gear:
- B2 and K1 activated. B2 same as first gear, so ratio is 1.58
- K1 locks up the front and central panetary sets, so the ratio is 1.
- Total ratio: 1 * 1.58 = 1.58


Fourth gear:
- K1 and K2 activated. Now all gear sets are locked up and the overall ratio is 1.

Reverse:
- B3 activated, this holds planet carrier P1 and thus the intermediate shaft and the outer gear O3 stationary.
- Front planetary set acts like a simple one, with sun gear S1 driving and outer gear O1 driven. Now the direction of rotation is reversed, with ratio 76/50 = 1.52 The resulting motion is transmitted through the hollow shaft connected to planet carrier P2, through the free wheeling clutch F, and through drum B2 to sun gear S3.
- the rear planetary set acts as a simple one, with sun gear S3 driving, and planet carrier S3 driven. Ratio is 1 + 76/44 = 2.73
- Total ratio is -1.52 * 2.73 = -4.15

Remarks:
- Basically the more modern MB transmissions are set up the same way, these use a Ravigneaux planetary set as the first compound gear set.
- The free wheeling unit F is NOT energized, that is why it is between brackets in Paul's list. It is however important in the all over scheme.

I am glad this is solved, since it bothered me enormously. After I verified the ratio of the compound set (with the brake band and vise setup) I was even more confused, because it proved my math was right (on the compound gear set). What I missed of course was the relevance of the hollow shaft and F, which I assumed was only used in Reverse.

Thanks again, Paul...

If any one has a copy for sale of the Tabellenbuch, I would like to buy one, does not matter in English or German.


Here a scetch of my interpretation of the actual set up of the members of the transmission:






1965 600 SWB #248
1968 6.3 #0347
1971 6.3 #5745 Euro
1979 6.9 #6857 Euro
1979 450SLC 5.0 EURO
1981 300SD
1989 560SEL
2003 CL600 Brabus T12 570HP

Edited by - paul-NL on 08/30/2017 12:33:10
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needamerc

Ireland
618 Posts

Posted - 03/25/2009 :  14:28:15  Show Profile  Reply with Quote
Albert. It seems to me that you are about ready to write 2009 edition Tabellenbuch yourself. Congratulations.
Eddie.
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