Re: Grapher 15.0.259 Torrent 2019 Download

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Julia Heaslet

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Jul 10, 2024, 10:37:50 PM7/10/24

to fichowsmitne

i have a question for YB24. I am new to grapher and am using it for a class. I have been trying to use the find root option in the equation menu and when I do it doesn't give me the x-value for when y=0. It keeps giving me numbers around it but it doesn't say y=0

I'm starting and studying Inventor and I could not understand some results of the Dynamic Simulation Output Grapher. Maybe the answer is simple but I did not find an answer in the help or on the Internet.

Grapher 15.0.259 Torrent 2019 Download

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Situation 1 - Here I understand the value of the forces, but if the forces 1 and 2 are equal, why there is a torque of 14 N.mm in the joint (Base Giratria, Torre) and if there is a torque, why "Base Giratria" ( Blue) does not it spin?

Force/Moment in joint is reaction Force/Moment in joint. As you know, joint is pair of two components constrained somehow together. Somehow means some DOFs are remaining, special relative movement is allowed (relative = first component relative to second), only some DOF remains. For instance Revolution joint has only 1 DOF - rotation around common axis. This is your example here with blue cup/disk.

Let's talk about Revolution joint. When joint is free, means no friction, idealistic, it means when first component rotates around Z axis, second component stays as it is, does not tend/want to rotate since joint is super lubricated/ideal. So there is no moment = no reaction/movement transfer from first (moving) component to second. "Relation between components is broken by lubricant" "Broken relation" = zero moment.

If you lock/fix Revolution (Lock dofs command) then moment can transfer from first component to second. "Broken relation is restored to tough relation" In this case you will get output grapher with your defined moment

Moment as vector has 3 compounds. Instead of graph of Moment M(which is SQRT (Mx^2+My^2+Mz^2)) be interested in Mx, My, Mz. You should get zeros in your sample, but your model is not correct, holes in blue cup/disk are not same, that's why there is non-zero moment around x-axis, y-axis (Mz = 0), byt total M 0. When I made blue cup/disk symetric I got M = Mx = My = Mz = 0

If you defined damping, it is something which explains moment in joint. Damping is resistibility against movement, gets stronger as relative movement is rising. So as blue cups rotates faster and faster, ability transfer movement from first component to second get stronger and moment convergences to total/given/defined moment

The Force/Moment of the joint shown in the output chart is the reaction of the Force/Moment applied to one of the two components of the joint measured in the other component of the joint that did not receive the Force/Moment applied. Right?

Not correct? why? We have an important point here, could you explain better? What model do you refer to? "Torque Testing.zip" or "Torque Testing 2.iam.zip" I think you refer to "Torque Testing.zip" and you are talking about its mass imbalance/asymmetry, right?

1. The Force/Moment of the joint shown in the output chart is the reaction of the Force/Moment applied to one of the two components of the joint measured in the other component of the joint that did not receive the Force/Moment applied. Right
I thought of reaction, but I thought of the reaction produced by inertia, so where does the reaction of inertia come in? reaction produced by inertia - yes, component wants to stay or move as it is moving, means change unwanted = 1st Newton's law. Only force can change state, force/moment can force component change state. Force/moment which changing component states goes through joint - force/reaction in joint shown in graph in this case

Your Mz = 0 since there is no friction/damping in your revolution joint, that is correct. But force 10 N (its plane) is 5 mm above joint so it creates/generates moment 10 x 5 = 50 N.mm. And you are looking into M which basically equals to Mx (since My, Mz are 0)

I liked it, very interesting to know about the manifestation of inertia! I could not see the model settings through the screenshot. Could you attach this example model (Assembly1) here for me to study it better?

I deduced that x is what has one arrow, y is the one with two arrows and z is the one with three arrows, but I'm not always right Smiley Sad and I can't fully trust my deductions, could you confirm that?

I use newer Inventor version than you, so you wont open assembly created by me. Nothing hard, just create beam with hole (or use square profile from Content Center), create assembly, place beam 3 times. Ground 1 beam, align second add Insert constrain for 3rd beam. Go to dynamic Simulation. Insert constrain will be converted to Revolution join, define imposed motion 90 deg/s and it is all.

Your issue presented in post 2 is the same as I showed in my post 1/Situation 1.
As I said in post 4, you do not provide enough information because you had not attached your model, but in the following post 5 I showed the solution to my situation 1 which also solved situations 2 and 3. Now I can say that I knew that it solves situations 2 and 3 too and I did not close the thread because of the doubts and your issue. If you had verified my post 5 solution in your model, you would have your solution at post 5.

As jan.priban explained in your model, I'm going to accept in this thread jan.priban's solution to your issue. You should at least give Kudos to jan.priban, people do not like to answer basic questions like ours.

Use the grapher to answer graphing questions. Plot your graphs on a grid and create different types of graphs using the palette of tools. Whether you use the small/large grapher, the drawing tool palette has the same icons and buttons. The displayed drawing tools depend on the type of graph you need to plot.