slimslowslider
VIP MEMBER
- Joined
- Jun 29, 2010
- Messages
- 317

I had to install +20 Emgo pistons in a 850 Mk2. I found out they were slightly heavier than the standard AE's (originals).
I wondered what this would mean to the balance factor. A friend of mine has a nice balancing jig, so I gave it a go.
Did some calcs first, had not done that before, and I really wonder if they make sense.
Can someone check them for me? Especially this compensating for oil I did.
The jig looks like in the pictures below. Thanks a lot, Bennie.
Balancing a 1974 850 Mk2 crankshaft with +20 EMGO pistons
Recip weight (Mrecip)
- piston, incl. rings/pin/circlip: 402 gr (original STD from original AE: 388 gr)
- conrod small end: 126 gr
Total: 528 x 2 = 1056 gr (STD: 1028 gr)
Rotat weight dry (Mrot,d)
- conrod big end: 318 gr
Total: 318 x2 = 636 gr
Rotat weight wet (Mrot,w)
- rotat weight dry: 636 gr
- measured volume sludge trap: 120 cc. Spec. weight oil: assume 0.85 kg/dm3
weight: 120 x 0,85 = 102 gr
Total: 738 gr
CALCS WET
Factory balance factor wet (BFw) is 52%. Bob weight then to be:
(BFw x Mrecip) +Mrot,w = (52/100 x 1056) + 738 = 1287 gr (STD: 1273)
CALCS DRY
Compensation to be added to bob weight for oil: 102 gr
Total bob weight: 1287 + 102 = 1389 gr (STD: 1375)
Corresponding balance factor dry (BFd):
(BFd/100 x Mrecip) +Mrot,d = 1287 >
BFd = ((1287 - Mrot,d) / Mrecip) x 100 = ((1287 - 636) / 1056) x 100 = 61.6 %. (STD: 62,0 %)
TEST DRY
Balancing the dry crankshaft in a balancing jig showed a bob weight of 1395 gr was needed.
Means an actual BFd of (1395 - 102 - 636) / 1056 = 62,2 %.
Note:
The Mk3 manual mentions a BF of 52% wet, and 63% dry, the latter with a bobweight of 1335 gr.
I had a old Mk3 crank, put it in the jig, I did not measure in detail but it showed it needed a little less bob weight than the Mk2 one. Don't know how much dirt is in the trap though.
I wondered what this would mean to the balance factor. A friend of mine has a nice balancing jig, so I gave it a go.
Did some calcs first, had not done that before, and I really wonder if they make sense.
Can someone check them for me? Especially this compensating for oil I did.
The jig looks like in the pictures below. Thanks a lot, Bennie.
Balancing a 1974 850 Mk2 crankshaft with +20 EMGO pistons
Recip weight (Mrecip)
- piston, incl. rings/pin/circlip: 402 gr (original STD from original AE: 388 gr)
- conrod small end: 126 gr
Total: 528 x 2 = 1056 gr (STD: 1028 gr)
Rotat weight dry (Mrot,d)
- conrod big end: 318 gr
Total: 318 x2 = 636 gr
Rotat weight wet (Mrot,w)
- rotat weight dry: 636 gr
- measured volume sludge trap: 120 cc. Spec. weight oil: assume 0.85 kg/dm3
weight: 120 x 0,85 = 102 gr
Total: 738 gr
CALCS WET
Factory balance factor wet (BFw) is 52%. Bob weight then to be:
(BFw x Mrecip) +Mrot,w = (52/100 x 1056) + 738 = 1287 gr (STD: 1273)
CALCS DRY
Compensation to be added to bob weight for oil: 102 gr
Total bob weight: 1287 + 102 = 1389 gr (STD: 1375)
Corresponding balance factor dry (BFd):
(BFd/100 x Mrecip) +Mrot,d = 1287 >
BFd = ((1287 - Mrot,d) / Mrecip) x 100 = ((1287 - 636) / 1056) x 100 = 61.6 %. (STD: 62,0 %)
TEST DRY
Balancing the dry crankshaft in a balancing jig showed a bob weight of 1395 gr was needed.
Means an actual BFd of (1395 - 102 - 636) / 1056 = 62,2 %.
Note:
The Mk3 manual mentions a BF of 52% wet, and 63% dry, the latter with a bobweight of 1335 gr.
I had a old Mk3 crank, put it in the jig, I did not measure in detail but it showed it needed a little less bob weight than the Mk2 one. Don't know how much dirt is in the trap though.