Recent content by Builder89

Sorry for the confusion. FR is not FA. FA is a vertical force applied from the ground on the wheel. FR = FS in the triangle model. (Force Right = Force Shock) Yes, 495 and 102 are the distances from right end to left end pivot and FM mount point to left end pivot. The purpose is to...

I just overlaid the free body diagram onto the bike for context. What is wrong with the post #46 diagram? It is the same as in the bike diagram above. You mean you think it has wrong dimensions? You lost me here. I have no idea what you're talking about past this point. "not...

Here is a rough representation of the context it goes in. All points on the triangle are pivots. FL is a rod attached to the frame on the bottom via another pin (pivot). The shock is only connected at the right point on the triangle and the frame up and left. Remember, we are doing the...

We calculated FS when the shock was connected directly to the swing arm. I explain in post #51 the problem that left me with as the method we used didn't apply to calculating the force on the linkage triangle when it was between the swing arm and shock, which is my final goal. Originally I...

JBA, as stated in post #53, we are analyzing the suspension without motion, sitting still, in equilibrium with the shock compressed 25% and need to calculate the spring rate at that point that allows FR to push said spring 25% (76mm). At this point, as I said in post #51, I'm unsure how to get...

Information about the shock isn't published in that level of detail. Basically you buy the shock and then you buy the spring for it that compresses 25% with the rider on the bike sitting still. The shock is built with this configuration method in mind. That's a starting point but rider...

OK then. I understand what you are doing here. It makes sense to me. I don't understand why the other one does NOT work but I DO understand what is going on here with the opposing torques being equal and the moments multiplied by distance from fulcrum. But this is simple with just the shock...

You're correct. The end goal is to get FS based on FR. But you asked how I would calculate the forces/torques and I told you. Apparently my method is wrong so I'm trying to understand why because the way you suggested, though cleaner in the case of the shock doesn't apply to my triangle case...

Guys. Maybe we aren't seeing eye to eye here. I'm trying to find the FM, which is a vertical force at this point in time where the swing arm is at 12° below the horizon. That FM is a number and that number should be able to be inserted as the up force on the next mechanism to sit on top of...

But guys, why is JBAs calculation using 495/102 and cos12 at all? I get that FM is the vertical force but that's the point. The vertical force was calculated so it could be applied to the next step of the mechanism. FM = 485.60N already accounts for the fact the swing arm is slanted 12°...

7.71FM? FS = 7.7FR above.

I'd like to believe you but I can't get it to reconcile with other information I think I understand so I must be wrong maybe somewhere else? FM was calculated above to be the vertical force down on that free body diagram. That makes me think there is mathematically an equal and opposite force...

Yeah, I understand. Your solution is simpler than the way I was trying to do it and it makes sense but I'd still like to know why my method yields a different result. My first attempt at FS I think was wrong, even for my intended approach. Let me explain. The shock is 50 degrees above...

Hmmm, really? I feel you're wrong. FL is the pivot point, O. FM and FR are torque forces working to twist around FL in opposite directions.

I'd love for you to look over them. :-) Here's how I calculate the vertical force up (FM) on the swing arm, which would then have the shock angle taken into account as applied to the bottom of the shock. The free body diagram would look like this: Let's just say FR = 100N (cause I forgot...