Theory Question

[moneal0514]

i am having a problem manually calculating a dimension for the geometry in a fully defined sketch. Theoretically if the sketch is fully defined there should be a way to manualy calculate all driven dimension values, correct?

[jletcher]

Can you be more clear, by what you are saying yes but sound to me you are having an issue.

Why are you having problems adding them?

[JDMather]

Correct.

Attach your file here.

You should not need to do any calculations - you can place driven (reference) dimensions.

[Cadmanto]

Are you talking about being able to calculate using existing dimensions to check your work persay?

Can you post an image with a real example of what you are exactly talking about?

If this solved your issue please mark this posting "Accept as Solution".

Or if you like something that was said and it was helpful, Kudos are appreciated. Thanks!!!! 

[moneal0514]

I am not having any problems with anything in inventor. My problem is trying to calculate lengths outside of inventor. The lengths i am trying to calculate are driven in my sketch, but i cant seem to manually calculate them. i am attaching a part that has a sketch of what i am trying to figure out. i am trying to manually calculate the dimension L

[moneal0514]

yes

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@Cadmanto wrote:

Are you talking about being able to calculate using existing dimensions to check your work persay?

Can you post an image with a real example of what you are exactly talking about?

If this solved your issue please mark this posting "Accept as Solution".

Or if you like something that was said and it was helpful, Kudos are appreciated. Thanks!!!! 

---

[jletcher]

Still lost what do you mean manually calculate them is not the dimension given what you want or are you looking for a formula to do something?

[Cadmanto]

Not exactly sure what you are trying to calculate, but in reference to the "L" dimension, it is coming off of the plane and there are no other stacked dimensions coming off of that plane to double check the "L" dimension.  Also seems like you are going to have to calculate the length of the arc and the distance to the arc's tangent point to the other end of the "L" dimension (opposite from the plane).

Can you explain in detail what it is exactly you are trying to calculate?

If this solved your issue please mark this posting "Accept as Solution".

Or if you like something that was said and it was helpful, Kudos are appreciated. Thanks!!!! 

[moneal0514]

i am trying to calculate this outside of inventor. pretend that i dont have inventor and i am trying to calculate the dimension L. I am just using inventor to lay it out to assist me with the calculation. I was just interested in knowing if it was possible to do the calculation from the given information, and since the sketch was fully defined i concluded that it had to be possible to do a hand calculation to find the length of L.

[moneal0514]

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@Anonymous wrote:

Correct.

Attach your file here.

You should not need to do any calculations - you can place driven (reference) dimensions.

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i guess this answers my question.

[jletcher]

Do you have a machinery's handbook? If so they show how to calculate this..

[Cadmanto]

I understand.  JD is right, the short answer is "Yes".  After all slide rules and pencils have been around a lot longer then Inventor or any other software's of this type.  Inventor just makes it easier to get these types of sketches to the final design stage quicker.  I think like James inferred to if you have a machinery's handbook that is a good place to start.

If this solved your issue please mark this posting "Accept as Solution".

Or if you like something that was said and it was helpful, Kudos are appreciated. Thanks!!!! 

[Curtis_Waguespack]

Hi moneal0514,

One thing to keep in mind is that it is possible to create a fully defined sketch in Inventor that you can not calculate just from the dimensional inputs, at least not without adding some driven dimensions. This is due to the use of sketch constraints that allow you create relationships between lines, without displaying numerical data.

Of course if you have the Inventor sketch available to you, you can simply add the driven dimensions you need. As an example here is a fully constrained sketched with only 3 driving dimensions. The rest of the inputs are sketch constraints. To take this thought to the extreme, we could create a fully constrained sketch with zero dimensional inputs by using the fix constraint.

It's a bit academic I suppose, but I thought I mention it since we're talking theory. In practice, I would agree with everyone else, you should be able calculate manually any fully defined sketch, but not necessarily any fully constrained sketch.

It's an interesting question, that I've bumped into in the real word when trying to document loads in Excel based calculations. Sometimes we just have to sketch a bit differntly to avoid automatic constaints and rely instead on dimensions, just so that the calculations can be easily verified.

I hope this helps.
Best of luck to you in all of your Inventor pursuits,
Curtis
http://inventortrenches.blogspot.com

[moneal0514]

i do as well as a degree in math, which is why this is giving me such bad heartburn,

From the attached pdf

1. The length of L is dependent on the length and geometry of the linkage

2. Points A and B are obviously dependent on the length of L

3. The x-coordinate of Point A is dependent on the geometry and length of the linkage, and thus the length L.

If i control the length of the linkage there is only one finite location for point A and length for L, however i always end up with more unknowns then equations. I could arrive at the solution by numerical analysis but i have been tasked with a simple "plug and chug" calculation.

What am i missing?

[moneal0514]

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@Curtis_Waguespack wrote:

Hi moneal0514,

One thing to keep in mind is that it is possible to create a fully defined sketch in Inventor that you can not calculate just from the dimensional inputs, at least not without adding some driven dimensions. This is due to the use of sketch constraints that allow you create relationships between lines, without displaying numerical data.

Of course if you have the Inventor sketch available to you, you can simply add the driven dimensions you need. As an example here is a fully constrained sketched with only 3 driving dimensions. The rest of the inputs are sketch constraints. To take this thought to the extreme, we could create a fully constrained sketch with zero dimensional inputs by using the fix constraint.

It's a bit academic I suppose, but I thought I mention it since we're talking theory. In practice, I would agree with everyone else, you should be able calculate manually any fully defined sketch, but not necessarily any fully constrained sketch.

It's an interesting question, that I've bumped into in the real word when trying to document loads in Excel based calculations. Sometimes we just have to sketch a bit differntly to avoid automatic constaints and rely instead on dimensions, just so that the calculations can be easily verified.

I hope this helps.
Best of luck to you in all of your Inventor pursuits,
Curtis
http://inventortrenches.blogspot.com

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i understand exactly what you are saying. i have actually tried to take into account the sketch constraints in my calculations. i guess i just missed a relationship somewhere.

[JDMather]

I recommend that you not override colors and use pink in your sketches as Inventor uses that color to indicate "sick" geometry.

[moneal0514]

normally i would never override the colors. i was going to use the colors to make it easier to explain, but i just didnt write it up.

[graemev]

Your sketch is good, but over-complicates the math required.  Mind you, it depends a bit on exactly what you're trying to solve for.  Are you varying the linkage and ladder-to-link-mount lengths to solve for a deployed incline angle (stowed in the vertical, obviously), or are you also fiddling with the location of the linkage attachment point on the ladder to match an existing deployment angle, or are you trying to find a relationship between some or all of the above?

Assuming the first option:  The link-ladder attachment geometry provides for a known angle B-Origin-C.  Lengths Origin-B and Origin-C describe two circles, as does link length A-B/C.  Controlling the "fore-set" of the link mount point (Ax)and height (L) should make the calculation of the deployment angle much more trivial.  (Though perhaps a bit above my pay grade, mathematically speaking.)