| oldfords | -- 11-13-2023 @ 6:56 AM |
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I have a question about the brake shoes on my 48 car. I am more familiar with the type of brakes that have an adjustment wheel at the bottom of the shoes. My 48 brakes have an eccentric new the top center of the shoes. When this is adjusted out so the brakes drag, what causes' the bottom halve of the shoe to contact the drum? When I activate the brake pedal, the cylinder pushes out the top of the shoe but it seems the bottom only pivots. This makes me think that only the top quarter of the shoe is all that ever contacts the drum. Please help me understand. al
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| juergen | -- 11-13-2023 @ 7:16 AM |
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In reality these Lockheed style brakes basically stop with the leading edge shoe. When the brakes are pushed out, the leading shoe is pushed down in the gap at the bottom by the spinning drum. This automatically adjusts the leading shoe to make contact on the bottom edge while the cylinder pushes out the top edge. In reverse, the "leading edge" is the rear shoe.
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| oldfords | -- 11-13-2023 @ 7:59 AM |
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OK That helps but it does not seem like a very good system but must work.
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| 51woodie | -- 11-13-2023 @ 10:53 AM |
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To quote the 1939-48 ford service manual: Anchor pins, )Ford & Mercury 1946-48 Self centering type brakes). On these models, anchor pins are fixed on backing plate and have special elongated washers which engage elongated holes in the brake shoe web. With this mounting, shoes tend to "center" themselves in drum when brakes are applied. It is important that the elongated washers are free to move in the shoe web. Adjustment is made from the back side of the backing plate. Turn each of the cam adjusters (see ref pics) until the shoe is tight, then back off until the drum just turns freely. Note: If the wheels are off, the drums must be secured on the hubs with nuts, when making adjustments. Sorry about the sideways pictures. This message was edited by 51woodie on 11-13-23 @ 10:55 AM
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