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#1
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Wally--
Do you have any old n/a 944 rotors handy? I've often wondered how close those hubs are to working with 20mm wide 944 rotors john
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johnL (aka H2OSB) '74 1303, Outlaw sedan (with a GL flavor) |
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#2
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You mean trying to put a rear 944 n/a rotor onto one of these alu front type 1 hubs?
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#3
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No, the front. It attaches the same way as the CB rotor. I've also considered getting a set of those hubs/rotors, removing the rotor and turning the hub slightly to fit inside a late 944 rotors (the ones that fit OVER the hub).
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johnL (aka H2OSB) '74 1303, Outlaw sedan (with a GL flavor) Last edited by H2OSB; February 25th 2018 at 15:57. |
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#4
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Ah! Now I see why you said used n/a rotor
![]() But euh...no, sorry. Due to me trying to minimize parts hoarding a bit, I brought all old rusty rotors to the dump.. sorry. |
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#5
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Quote:
I actually kinda got on the guy at CB who designed those hubs/rotors. I asked "why would you re-invent the wheel, so to speak, when Porsche rotors are readily available?" He had no answer.
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johnL (aka H2OSB) '74 1303, Outlaw sedan (with a GL flavor) |
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#6
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Quote:
I think all porsche rotors (except the 914 rears) are vented and they wanted something for the type 1 market? I'll try to measure the bolt spacing ;-) |
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#7
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Their source for un-drilled rotors has dried up so they needed something to offer a 5x130 pcd for the front. A lot of potential...which I don't think they realized (well, perhaps not until I started bugging them about it)
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johnL (aka H2OSB) '74 1303, Outlaw sedan (with a GL flavor) |
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#8
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Anyone mind going over the math for why a stepped MC does not effect the preasure output by each cylinder?
A dual MC is set up with a floating piston between the two separate cylinders. ![]() if we assume we are at steady state (piston no longer moving, just generating more or less force), that the brake line ports are plugged, and that this piston is not connected by a bar and the spring force is minimal when breaking, the pressure in the primary circuit (P1) is going to be the input force (Fin) divided by the plunger area (A1). Without any solid connection or added force between the two channels, Fin must be equal to the force the fluid in the first circuit applies to the floating piston (F1) must be equal to Fin. Secondly, the force the floating piston applies to the second circuit (F2) must be equal to the force applied by the fluid in the primary circuit. This force (F2) is equal to the pressure (P2) in the piston multiplied by the area applying the pressure (A2). Therefor, if the area of the two circuits are not identical, the pressure within them will be different as well
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