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it's more complicated
And that's before you even throw in the preloads and components which are not fully kinematically constrained.
gbj_tester-
Isn't it more the case that because there are a lot of parts that are kinematically constrained (though in complicated ways), the entire system (once you take into account the spokes) becomes massively overdetermined, which is what makes the exact load calculations so difficult?
But then also when you want to get into the details, you need to take into account second order effects because it's not actually a rigid body system, rims and spokes etc. can flex as well.
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you can make it more complicated but the result still is disc brakes put more load on the the spokes than rim brakes do on average. The exact ammount will depend on a number of factors but it is wrong to assume they don't.
SwissChap
cycleclinic
Yeah this situation is a bit more complicated than just a single equation. If you actually want to understand all the forces in play, you need to draw a free body diagram of the whole situation - and the specific force acting on one spoke is not that easy to determine even then. That's with making the assumption that a spoke can only take tension and no compression: otherwise it's more complicated still.