[2thumbsup] :-D
Yes zekeman I believe you've cracked it!
It is good job my cam angles are < 45 degrees and not close to 90 degrees or some other physics would kick in.
Tigers drive and my Sunday best do hit close 80 degrees and they sound good. There's a clue there but I won't go down that road...
Hi folks. [sadeyes]
Here is my take on the sphere/pendulum wedge impact.
See attached jpeg.
Fig 1 shows the relative velocity of sphere and wedge.
For convenience consider the sphere hitting the wedge not the other way round.
Fig 2 shows the freebody deflection of the sphere.
My contention is...
Thanks zekeman.
The camera shows the motion of the sphere to be greater than the surface height of the cam (when the leaf spring has a mass on the free end). Subsequent impacts buld up!
As a golfer I am well aware of what happens when a sloping cam hits a stationary sphere. I am not very good...
Hi desertfox and GregLocock
Many thanks for staying with it. I am diverted onto something else for now but will be back.
GregLock we have more in common than you know.
In the 80's I was modelling 1/4 car suspension with humble computers. I wish I had todays toys. Tyres were difficult.
Every...
[surprise]The problem expands!
I am delighted that you guys are trying to help but I can't be completely clear on what is going on, so I have attached an equivalent (well a rough equivalent).
The file 'CatchCam.jpg' has been uploaded.
The notched cam travels right to left, pulled by a heavy...
:-D Ahay now we may getting somewhere.
I like the energy balance because I have not disclosed the real situation. It gets worse!
The small sphere is actually a bend on the end of a leaf spring. The bend is an arc of a circle. The spring is anchored at the opposite end.
The wedge is a cam...
[ponder]Reminder: I am trying to determine the initial horizontal velocity of the sphere. It is a collision. The bodies are hard and elastic. The sphere does leave the wedge, I have high speed video to prove it.
I saw the problem as a collision between objects.
Impulse forces, duration of...
hello zekeman
Just returned to the problem and looked at using Vx = VyTan(Q) as you suggested.
:-( This can't be right! Vy.Tan(Q) is the vertical velocity to gently run up the wedge surface. No impact!!
Velocity due to impact must be higher than this. The sphere has to bounce off the wedge.
:-D Many thanks zekeman.
I can now calculate the restoring force to return the pendulum to the wedge in a given time.
Incidently, on the second impact I will add on a horizontal velocity of Cr.Vx2 where Vx2 is the velocity when the pendulum returns to the new part of the wedge. Cr is the...
I have a massive wedge rising vertically and impacting on the side of a pendulum.
The pendulum consists of a hard sphere suspended from a rod and is constrained to only rotate about its’ fulcrum.
The wedge nose/slope has an angle Q to vertical.
What is the instantaneous velocity Vx (horizontal)...
I am running an Excel VBA programme where the VBA code calls the Solver and moves on immediately without waiting on the Solver. In fact the Solver does not run! :-(
When running in the development environment it works!
When starting the code via an event such as command button press, it fails...
Disclaimer I believe the expressions are correct but use at your own risk.
T1 is the general expression for driving a variable inertia load.
T2 is a particular expression for controlling the variable inertia AND using a constant speed shaft.
Torque for variable inertia load
Torque T1 =...
Eureka! It seems everyone is correct.
The momentum equation applies.
What was declared as 'Drive torque for the variable inertia' in the solid model, is in fact the 'System torque' as it includes the control means for varying the inertia.
In this case a set of cams was used to vary the radius of...
Many thanks prex.
Yes everyone agrees that the momentum approach is correct.
Rather than blame Pro-mechanica, I will go to my colleague and see how he set his solid model up.
There may be an assumption in his model that it is a conservation of energy system.
I had included in my model...
Hello quark.
Think of a governor for steam engine where two balls are mounted on opposite corners of a parallelogram and the central rotating shaft goes through the other two corners.
The inertia can be varied by contracting or extending the distance between the two balls. The radius of rotation...
Thanks to zekeman and sreid.
Ok my second method listed is Newtons Second Law applied to rotating inertia and this was my first solution when I started the model.
However we have a solid modelling package which can also determine torque for a variable inertia provided the inertia is strictly...
Hello quark
If you mean should there be any DJ/dq then I can only point out that both equations for T use dJ/dq and that the first equation agrees with our solid modeller software.
If you mean that one of the maths steps involving dJ/dq is wrong can you elaborate please?
I have two expressions for torque with variable inertia. Will someone check out the maths for me please?
Deriving Torque from energy change
Work T.dq = Energy change = d(1/2 w2.J ) q = rads, w = rad/S
d(1/2 w2.J ) d( w2.J )
Torque T =...
I am trying to use the text parameter of one shape to determine a number and use this in a shapesheet formula for another shape.
In particular I am using VB to write a string (number) to the text parameter of one shape.
The shapesheet for another shape should be able to use the numeric value of...
More info - but still no easy solution.
When I load Excel from VB the default Autoshape units are inches. No Regional settings.
When I load Excel from the desktop it first loads my Personal.xls which has regional settings and so uses cm for the Autoshape size.
