Advice on how to figure out a dimension

[9fly]

This isn't AutoCAD specific, just more of an exercises to get me re-acquainted with AutoCAD as I haven't used it in a while. On this drawing, what's the best way to interpret the dimension that has the 60 degree feature? It looks like one of the 60° arrows in the drawing is pointing to a hidden line? Or maybe no, I am not sure that's what I have the question mark there. I was hoping to simply figure out where that line started on the base plate and I could simply draw a 60° line and be done but I am not seeing how. Maybe i am just not thinking this morning though.  Regardless, where this was done in AutoCAD or Inventor I'd still be stuck. Any help/advice would be appreciated...again, it's just an exercise to get me back into using AutoCAD...nothing too important! 

 

 

[guran]

To me, the angle shows how much the pipe section is inclined from a vertical line.

[nukecad]

That's also how I would interpret it - 60 deg. from vertical.

 

I'd probably have dimensioned it at the other end as 30 deg from the 1.25 dimension leader on the centreline. But TBH that's all horrible dimensioning. There is nothing to position the boss (other than being on centre) and the 2.5 goes to the centre of a lug/hole you can't even see and have to infer from the '2x .38 THRU'.
EDIT - I can see the lug/hole in hidden detail now that I'm on another device.

 

When dimensioning anything you should think of how the person making/inspecting it will be able to measure it.

Edited October 25, 2020 by nukecad

[9fly]

You mean the pipe section is inclined at 60°, right? I wonder, when creating the front view, how far would I go into that base plate before drawing the 60° line?  ....I know the center-line of the pipe  is at 1.0" if you are looking at it from a top view, but on the front view from the left of the base plate how far into it would I go. If the answer is obvious, I'm sorry I didn't get much sleep last night lol 

 

Also, just random though.....one thing about AutoCAD...I noticed that you use geometry a bit more than in Inventor...something I could use improvement on for sure. So far, I am enjoying using AutoCAD after being away from it for a pretty good while.  Thanks all.

Edited October 22, 2020 by 9fly

[steven-g]

I would drop a vertical pipe down first centered on the base plate and then using that same centre point rotate the pipe 60° and then 'push' it through the base before cutting a bit off the bottom.

[9fly]

Okay, I think I understand. I never thought about doing it that way but that sounds like it would totally work. Thank you!

[eldon]

It would seem curious to me that only one given dimension is to four decimal places. The rest of the dimensions seem to be much more run of the mill.

I would draw out what you know and see if it gave some more rounded lengths.

[9fly]

Thank you and you're right, that is odd...shouldn't ignore that...definitely will do it...

[SLW210]

The 60° is the angle of the tubing. The decimal places just indicate how accurate the part needs to be, many places I have worked had a general tolerance chart, if the dimension was in fractions, it listed for 1/64", 1/32", 1/16" if decimals the number of places after the decimals. 4 places seems unnecessary, though it is just a sample.

 

Like This

[nukecad]

On 23/10/2020 at 12:10, SLW210 said:

 4 places seems unnecessary,

 

Depends on what you are doing.

 

For hydraulic seal housings, bearings, injection moulds, etc. a tolerance of less than a thousandth of an inch (or less than a hundredth of a millimeter) is not unusual.

 

In the sketch above though it has simply been calculated from a set length (don't have CAD here but it's 1.625" if my triganometry brain is still working) at the Cl of the boss when cut through at the angle.
EDIT That is wrong, my trigonomatry brain obviously wasn't working. See below.
It probably doesn't need to be that accurate in real life, it's just a calculated number.
(Bad dimensioning again - if you specify something that accurately on the drawing then it has to be inspected to that accuracy, which puts the manufacturing cost up unecessarily).

 

PS. I also note that there is no indication of whether the hole continues through the base plate or not. It could just be a blind boss welded  onto a solid base plate.

Edited October 25, 2020 by nukecad

[eldon]

I would have thought that in the context of a drawing exercise, that four decimal places would be highly significant.

[nukecad]

Nope, as said in my post above it's just been calculated from 1.625" at the centre line. (Wrong, see below).

 

There is no way it could be critical in a part like that, especially given the nature of the other dimensions.

2.815 or 2.82 (or even just 2.8) would probably be sufficent in real life.

 

But yes as a drawing excercise then the student should do what is there.

Edited October 25, 2020 by nukecad

[nukecad]

It's always been a bit of a funny thing with imperial dimensions.

 

1.625" looks as if it should be more accurate than 1-5/8", but they are exactly the same.

[eldon]

I think you are thinking like a seasoned professional.

If you take the tube and rotate it about its centre line and the top face of the base. Then take a slice from the top edge of the base plate perpendicular to the tube. Then what is the length of the tube? 2.8146. Any more or any less, the tube would not be in line with the top edge of the base.

It is a drawing exercise!

 

 

 

 

 

[eldon]

Now I am back at my computer, here is a picture to show what I mean.

 

 

[nukecad]

Yes, now draw in the tube centre line and measure it from the tube end face to where it intersects the angled face, I believe that you will find it's 1.625" or 1-5/8" if you perfer. (EDIT, that's wrong see below).

 

As you rightly say it's a drawing excercise, and 2.8146 would be the total length of tube needed.
However should they be teaching students bad dimensioning practice? (Even in just a sketch).

 

PS. I reckon that component is a simple bracket for holding a flagpole, awning support, or similar at 30 deg to a wall, (although the tube may be a bit short for that), anyone else got a guess?

Edited October 25, 2020 by nukecad

[eldon]

1 hour ago, nukecad said:

Yes, now draw in the tube centre line and measure it from the tube end face to where it intersects the angled face, I believe that you will find it's 1.625" or 1-5/8" if you prefer.......

 

 

You get no argument from me on further drawing, but I was trying to limit my response to the OP's original request, and point out an anomaly in given data could lead to a solution - ideal for the academe.

[nukecad]

I agree the excercise may have been originally intended to see how a student will cope with unclear information on a sketch.

(And work out for themselves where that 2.8146 comes from?).

 

Unfortunately sometimes these excercise sketches/drawings get copied by others and the original intention, along with the tutor feedback, gets lost.

It's a particular issue with online courses where there may not even be a tutor to give feedback.

Which is one reason why students have to come to forums like this for help.

 

At least this got my brain refreshed a bit with inches. It's been almost 40 years since I worked in anything but metric.

(TBH I'm continually surprised the US still uses Imperial dimensions - you left the Empire a couple of hundred years ago and your coinage is decimal/metric).

Edited October 24, 2020 by nukecad

[steven-g]

4 hours ago, nukecad said:

Yes, now draw in the tube centre line and measure it from the tube end face to where it intersects the angled face, I believe that you will find it's 1.625" or 1-5/8" if you perfer.

Have you tried???

[eldon]

18 hours ago, nukecad said:

Yes, now draw in the tube centre line and measure it from the tube end face to where it intersects the angled face, I believe that you will find it's 1.625" or 1-5/8" if you prefer.

......................................

 

Do you mean something like this? I can't see any 1.625. Perhaps you could post a picture to enlighten me. Thank you.