Installation & update
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The great divide

Anonymous
Not applicable
Hi Folks,

As a very new ArchiCAD 9 user, I'd be grateful for a little assistance.

To try to get up to speed with things, I'm rattling through the Training Guide that came with our version of ArchiCAD 9, which involves drawing the Drucker-Brownstein residence.

Unfortunately, I'm having problem inserting some windows.

As part of the training guide, you are required to insert 4 windows on an equal spacing around the perimeter of an arc. However, selecting the division option does not give the correct result (the wall width at the ends of the wall are far greater than the gaps between the windows).

To try and resolve the problem, I've opened the next file in the sequence of training files where the windows appear in the correct locations; however, when I compare the location of the windows with the snap points, the windows locations do not agree with the snap points.

So, come on then guys, what's going on?

Is it me, or are you all rolling abount on the floor saying "he fell for that old nut shell"

I should also say, that I may be a new ArchiCAD user, but I'm an old, old hand when it comes to AutoCAD (if I'm allowed to mention that name in these parts), and with the exception of this little glitch, I'm seriously impressed so far.

Many thanks

John
19 REPLIES 19
Anonymous
Not applicable
I very very rarely have designs with curved walls, but i will take a "stab" at this one. This may look like trouble but it was not too bad. I am to lazy (and busy) right now to explain this in words though.


Did I mention that I almost never do curved walls. Too much of an engineer for the fancy curves.
Anonymous
Not applicable
Peter wrote:
Hello Matthew,
It occured to me to try the multiply/distribute command to
array hotspots or windows along the edge of a curved wall,
but it doesn't work.
Meaning, it does not put the hotspots in the right place
for the present purpose.
Could you explain how this is done.
Thank you,
Peter Devlin
Don't do it with hotspots. You need to use the element itself or something that matches it in width. Place one (to be removed later) at the end of the wall (off the end actually - just touching at one point). Multiply with the distribute minus one option dragging the outside corner of the element (the one not touching) to the far end of the wall (or space) to be divided.

I hope this makes sense. It is much easier to do than to describe, but I am too tired right now to make any pretty pictures.
Anonymous
Not applicable
Matthew wrote:
...It is much easier to do than to describe, but I am too tired right now to make any pretty pictures.
Hey, no slamming the Pretty Pictures. That is my "Bread-and-butter" here. I always have to attach a picture, because, one: I can not type very well and Two: I are a engineer and words don't come to easy for me.

No attached picture this time. Hey I am getting the hang of this....

Jay
Anonymous
Not applicable
Hello Matthew,
I tried your method and, amazingly, it works !!
Thank you for teaching me this.
For the general case one has to do some careful setup
but it is quick to do.

Thank you,
Peter Devlin
Anonymous
Not applicable
After learning how to distribute the windows using Matthew's method,
it occurred to me that one could get the divisions method to
work by stretching one end of the curved wall exactly one
window width longer and then setting divisions to five.
This will then show hotspots at one corner of each window.
After placing the windows one stretches the wall back to it's original length.

So the division method does work after a fashion using this "trick".
Thanks to Matthew we are all spared the horrible arithmetic
I had suggested in my previous post.

whew !
Peter Devlin
Anonymous
Not applicable
Peter wrote:
Hello Matthew,
I tried your method and, amazingly, it works !!
Thank you for teaching me this.
For the general case one has to do some careful setup
but it is quick to do.

Thank you,
Peter Devlin
It is my pleasure. Glad to be of service.

I find this trick to be useful in all sorts of circumstances, such as laying out muntin bars in custom doors and windows.
Anonymous
Not applicable
Peter wrote:
After learning how to distribute the windows using Matthew's method, it occurred to me that one could get the divisions method to work by stretching one end of the curved wall exactly one
window width longer and then setting divisions to five. This will then show hotspots at one corner of each window. After placing the windows one stretches the wall back to it's original length. .........
After "getting-my-head-around" Matthew's method (where is a nice picture when you need one ), I now do not see a need for the Divide Hotspots.

Because I was using a single Arc as the wall, (not read practical since there will be most likely other walls attached to this wall to enclose the space) I could not place the first window, which is to be removed later, just touching on one end.

After placing a Hotspot on the Original end location of the wall, I just stretched the wall at least as far as the window dimension or a new wall can be used with the same Radius (no Calculation neccesary). Placed the side of the temporary window at hot spot and sent it into the temporary stretched portion of wall. From there I use the "Matthew Method". Delete temporary window, stretch (squeeze) back the wall to original size, or if a new wall was placed just delete the wall and the temporarily placed window is deleted also.

Not that difficult and no Calculations. Thank you Matthew.

Am I making this more difficult by adding the temporary wall? I could not place the temporary window otherwise.
Anonymous
Not applicable
Jay,
I think you have answered your own question
I did want to ask you about something in your
post of May 26.
You show a formula in your "pretty picture"
that reads arc=(05*180/PI*15)
This looks similar to the arc length formula
but where does the 05 term come from ?
Thanks,
Peter Devlin
Anonymous
Not applicable
Because this is a low pixel count JPG, the decimal point was "non-existent". The number was 0.5 (6" = 0.5 feet) for the arc distance. But this mental routine need not be performed. The formula is helpful though for finding the required angular distance for a given arc and Radius, as you know.
Anonymous
Not applicable
Thanks Jay,
I really was staring at that formula and wondering
what I was missing.
I thought it might be 0.5 but I could not see any evidence
of a decimal point.
Peter Devlin