i know this is a dumb question but can you run 42-8 drums on 37-41 spindles and have the brakes still work properly.... i know the later brakes have a wider shoe but will there be a problem ..reason is i have all new bakes for round back spindles but the drums are bad and i have some later brake drums .. thanks for any info kyle
Yes, with no trouble at all. Of course you have to use the hub/drum assembly rather than just the drums.
The shoes are not wider as far as I know. The lining kits are the same for all years 39-48. The later shoes self adjust at the pivot end, the older ones have to be adjusted on an eccentric. As stated above, any 39-40 backplate/ shoe assembly will fit the spindle, and any 39-48 hub/drum assembly will also fit, regardless of the backplate used. Mart.
thanks everyone for your speedy replys . only reason i ask was cause when i ordered the new shoes this time the guy at job lot was saying one shoe was wider but i dident thinks so cause ive done this before .. figured id see what che commity said .. thanks once again .
You can use either year drum on any of the early ford spindles since the all use the same bearings. The difference is the early drums have the hubs behind them and the 46-48 drums the hubs are on the outside making them popular for buick drum conversion. 39-48 brake shoes are all 1 3/4 inches wide but have different shapes on the bottoms for the lower cams. 39-41 are round and d-shaped in the middle. 46-48 have flats on the sides and round in the middle. I prefer the early year brakes meaning 39-41 since you have more shoe adjustment up and down plus toe. Kevin
how about 49 - 50 ford car brakes (backing plates, hubs & drums) on the early (37-41 or 42-48) spindles?
Never heard of anybody doing that swap so there must be a basic compatibility issue. (IE they don't fit!) People tend to use the f100 stuff (or lincolns) for self energising 11"ers. They are more of a bolt on (I guess) than the car stuff. Plus the car hubs have the wrong bolt pattern .. 5 on 4.5" rather than 5 on 5.5". Mart.