Starting force of a falling object on a compound lever

[Builder89]

FS cos(52) (102)=FR cos(12) (495).
So FS=7.7 FR.

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.

How do I do the same thing for the linkage triangle in post 46? I'm not sure where to start to get FM.

I'd use torques to get a relationship between FL and FR.

FL(sin80)(15) = FR(sin70)(30)
FL(14.77) = FR(28.19)
FL = 1.91(FR)

Then I'd use sum of forces to get FR:
0 = FM - FLV - FRV
FM = FLV + FRV
FM = FL(cos25) + FR(cos55)
FM = 1.91(FR)(cos25) + (FR)(cos55)
FM = 1.73(FR) + (.57)(FR)
FM = 2.30(FR)
FR = FM/2.30

But how to get FM since I can't take the vertical component from the swing arm calculation. Do I do a torque calculation somehow using both the 15 and 30 sides of the triangle together like FS was used in your quote at the top of this post? I'm lost here.

 

[haruspex]

How do I do the same thing for the linkage triangle in post 46? I'm not sure where to start to get FM.

In the drop scenario, you cannot compute F_M independently of the shock behaviour.
First, the forces vary during the landing, so you have to decide what force you want to know. I suggest the peak force - ok?
The graph of force against time (whether it's F_M, F_S or F_R) will depend on how the shock reacts. The more compression it allows, the longer the time, so the lower the forces.
Just how much do you know about the absorber? In post #22 I listed some possible features and parameters.

 

[Builder89]

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 preference and terrain could result in the rider choosing a different spring for the occasion. Let's take advantage of that configuration method.

At this point in time represented by the diagrams, I have a load of 100kg (980N) from the rider and bike just sitting there. The swing arm is at this 12° location without movement. At this point, the shock is compressed 25%. The total possible travel (distance from fully uncompressed to fully compressed) of the shock is 3in (76mm) so this 980N at the axle is compressing the shock 19mm right now. The measurements in the diagrams are mm.

We should be able to calculate the force pushing on the bottom of the shock at this time. Then we can calculate the spring rate that would allow the spring on the shock to be compressed 19mm by this force at this time.

Once we have the spring rate, we could calculate the other forces, such as the force it would take to compress the spring 100% and how much force would be applied to the frame, etc.

 

[JBA]

The spring and the shock are going to act in concert to absorb the mgh energy of the falling mass so in order to calculate the total shock/spring travel and the final force on the shock/spring connection points it is necessary to know how much energy the shock absorbs per unit distance of travel.

i.e. mgh = 1/2*k*x^2 + (shock E/dx) * x

If you can determine the energy absorbed by the shock at a desired x travel, then that x value can be plugged into the spring section in that formula to solve for the required k (spring rate) for the spring.

 

[Builder89]

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 the FM going into the bottom of the triangle and I'll need that in order to calculate FR.

 

[JBA]

Your first post #3 bicycle figure best illustrates the force vectors associated with your spring mounting; and based upon that figure and succeeding angles and connection dimensions for the swing arm and shock/spring and FR load points, it appears to me that we have already done that by calculating FS = 7.71 FR, from that k*x = 7.71 FR or k = 7.71 FR / x so if you want the spring compression to be 3" then k =7.71 FR / 3 (lb/in).

Edit: or the appropriate mks value that is std for springs.

 

[Builder89]

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 left the triangle out because I figured I could calculate FM from post #3 and apply it directly as FM vertical component at the bottom of the triangle. But apparently I was wrong about that.

You can see in post #51 how I intended to use the vertical component of FM from post #3. But instead we got FS, which doesn't apply to the bottom of the triangle.

 

[JBA]

OK, at this point it would help me understand what you need if you would just provide a bicycle figure like that in post #3 with the added linkage you want to use. All of these partial linkage diagrams you have given do not allow me to see the context of the complete structure.

 

[Builder89]

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 calculation in equilibrium at this point with the shock compressed 76mm. As such, this is how it looks. All numbers are mm.

And, as stated before, the FA is just the portion of weight on the bike that is at the rear tire.

 

[haruspex]

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 calculation in equilibrium at this point with the shock compressed 76mm. As such, this is how it looks. All numbers are mm.

And, as stated before, the FA is just the portion of weight on the bike that is at the rear tire.

I believe your force diagram for the linkage in post #46 is wrong. (If correct it would be overspecified, with two unknowns, the magnitudes of the two forces, but three equations.)
In your diagram just above, which of the connections in the linkage are free joints? Even if they are all free joints, the F_R force is not constrained to act along the rod that goes to the rear wheel. That is a rigid rod, so it can generate a restraining force in any direction.
If we go back to your post #46 diagram, we can make the angle you show as 110 degrees an unlnown. That gives us the right number of unknowns. From here, it is just a matter of writing out the horizontal and vertical force balance equations (ΣF=0) and the torque balance equation. The easiest is probably to use the joint with the unknown angle as axis for torque. That allows us to determine F_L immediately.

(It is unfortunate you have reused "F_R". This might lead to confusion.)

 

[Builder89]

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?

which of the connections in the linkage are free joints?

All points on the triangle are pivots.

Even if they are all free joints, the FR force is not constrained to act along the rod that goes to the rear wheel

You lost me here. I have no idea what you're talking about past this point. "not constrained to act along the rod"? generate force in "any direction"? make "110 degrees as unknown"?

FR confusion? FR is still the same, the force the right point of the triangle feels in the down,right direction.

 

[haruspex]

I think I simply don't understand your diagram in post #59. The way the load is transferred to the shock looks crazy. It couldn't work.
The confusion over F_R comes from the diagram in post #33, where it seems to have been used in place of F_A.

 

[JBA]

Is the overall length of the swing arm still 495 mm and the distance from the swing arm front pivot to FM 102 mm?

Just for my education, what is the primary reason/purpose for this linkage system?

 

[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 multiply the distance the shock is compressed. The right end of the triangle will move further than the FM mount point. The dimensions of the linkage can also be manipulated to create a non linear force on the shock through the wheel travel... for example becoming more stiff as approaching 100% compression or other patterns.

haruspex, I'm not sure which part is looking like it doesn't work. wheel pushes up on swing arm which rotates about the attach point on the left end. It pushes up on the FM mount point which pushes the triangle up. The triangle left side is held down by the linkage bar attached to it. The right side of the triangle rotates up and left as the bottom is pushed up and the left side moves more slightly up and to the left.

 

[haruspex]

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 multiply the distance the shock is compressed. The right end of the triangle will move further than the FM mount point. The dimensions of the linkage can also be manipulated to create a non linear force on the shock through the wheel travel... for example becoming more stiff as approaching 100% compression or other patterns.

haruspex, I'm not sure which part is looking like it doesn't work. wheel pushes up on swing arm which rotates about the attach point on the left end. It pushes up on the FM mount point which pushes the triangle up. The triangle left side is held down by the linkage bar attached to it. The right side of the triangle rotates up and left as the bottom is pushed up and the left side moves more slightly up and to the left.

I don't think I am going to understand that latest diagram unless you indicate on it which intersections of lines are actual joints (rather than one member passing behind another) and which of the joints are pivots. While you are at it, please assign a label to each joint to aid discussion.

 

[JBA]

One general comment on your plan to " lengthen the stroke of the shock" since the shock and spring are operating in parallel you should realize that the force that you lever applies to the spring for a given FM will by reduced in proportion to the amount that you increase the lever length (if the lever length is doubled relative to the wheel travel; then, the force it applies to the spring/shock connection is 1/2 of the that of the shorter lever); and as a result the spring compression length for a given load will also reduce proportionally; and you will find that there is no actual additional shock/spring travel provided by your lever mechanism at all.