Compound Angle problem

[JAJIII]

Hello All,
This is my first post and I hope that I am putting this problem in the correct area. I am a cabinet professional that uses a 3D solid modeling program called Solid made by Planit Solutions. I am trying to build a hood cabinet (see picture below) but I am having problems with the compound angles. If you look at the picture below the 0,0,0 point of the cabinet is at the back, bottom, left corner. The X-Axis runs from Left-Right, the Y-Axis is Bottom-Top, and the Z-Axis is Back-Front. I am trying to figure out how to figure the length and x/y/z placement coordinates for the front pieces that are on a 45 degree angle. In the example below I have put in some of the equations that I am using for the left, right, and front of the part. But I can't use those equations because of the compound angle. Can anyone give me some help with this? Here is a drawing of what I am trying to accomplish:

Any help would be greatly appreciated. I have problems like this that come up often and I am looking forward to using this resource for my answers. I also look forward to answering some of the questions and contributing the more I learn. Thank you so much for spending the time to help other,

John Jones III

 

[galactus]

I hope I am understanding you correctly.

You can find the length of side C by using Pythagoras:

\(\displaystyle \sqrt{25^{2}+12^{2}}=\sqrt{769}\approx{27.73}\)

known: We have side a=7, side c=27.73, angle B=64.4.

Now, we can use the law of cosines to find the side opposite from side C.

Call it side b:

\(\displaystyle b=\sqrt{7^{2}+(\sqrt{769})^{2}-2(7)(\sqrt{769})\cos(64.4)}=25.5\)

We can use the law of sines to find the other angles:

\(\displaystyle \frac{a}{sin(A)}=\frac{b}{\sin(64.4)}\)

\(\displaystyle \frac{7}{\sin(A)}=\frac{25.5}{\sin(64.4)}\)

\(\displaystyle \frac{7\sin(64.4)}{25.5}=0.247561869994\)

\(\displaystyle \sin^{-1}(0.247561869994)=14.333 \;\ degrees\)

The other angle, angle C, is \(\displaystyle 180-(14.333+64.4)=101.27 \;\ degrees\)

Is this what you were getting at?.

 

[stapel]

I am trying to build a hood cabinet (see picture below)...

Would it be possible for your to post the image for this exercise somewhere that doesn't require us to guess the password for your secure FTP server?

Thank you.

Eliz.

 

[galactus]

Stapel, are you referring to the picture?. I can see it OK.

 

[stapel]

Stapel, are you referring to the picture?. I can see it OK.

The link starts out as "ftp://cvforum@wan.planitsolutions.com/cvforum...", and I get a pop-up box asking me for the password.

I don't usually surf in Internet Explorer, and a quick check confirms that IE does some sort of end-run around the security for the "Plan It Solutions" forum. :shock:

Note to JAJIII: It looks as though, in IE, I can open your account...?

Eliz.

 

[JAJIII]

There is nothing there of any interest to anyone except a cabinet vision user. Also my account only lets me put stuff up there and download it. You can't delete or do anything else. That is the only place that I know of to put my pictures. I have an ftp setup here at my house but I can't get passive mode to work so you would have to turn off passive mode to be able to open anything.

Galactus,
I thank you for the help, but that was not what I needed. What you have put there doesn't work. If you notice the top is shorter than the bottom so it sort of looks like this:

| 12 |
__________ __
| *
| *
| *
| * 27.73
| *
| *
------------------- ---
| 17 |

and it is rotated at an angle. So actually you would have to find the side of the triangle that I think you identified first as being 27.73 would actually be
sqrt((27.73*27.73)+(5*5)) = 28.177. That would be the left part of the triangle. That is as far as you can go. I guess my picture wasn't clear but the 64.4 is the angle that the left end, right end, and front are angled at for it to meet with the bottom of the top rectangle. The illustration above is the left/right ends. The front is similarly shaped but it is anlged on both ends. Like this

| 30 |
. ___________________ __
. * *
. * *
. * * 27.73
. * *
. * *
. * *
. -------------------------------------------- ----

. | 40 |

With the parts angled on the ends and the parts rotated at an angle it changes the way you can work the problem.

For me to figure the length of the clipped part I have to first figure out the bottom length of the bottom of the triangle. Now it could be figured as 7/sin(45) because that part will always be a 7x7 clip and we all know that using the law of sines we can figure that third side. So now I know that the bottom of that clipped triangle is 9.8995. Ok we know that but and we know that the triangle we are working with has equal sides because both the front and side edges will be 28.177. Now that we know that It is possible for me to figure out the height of the triangle by bisecting it and using C^2-B^2=A^2. So now we have the height of the triangle piece is 27.73 that is on the corner. BUT I still don't know the angle that I need to rotate it at to make it meet the bottom of the top rectangle. After talking my way to here I have come to the conclustion that it is the same angle as the front and sides because the height of the triangle is the same height of the sides and front. Thanks for all your help though. If I hadn't had to do this explination I would have never figured it out. Thanks again,

John Jones

 

[pka]

I cannot see it with Netscape.

 

[jwpaine]

FTP needs to die.

If you really need ftp for linking photos...setup anon access and include the anon/ftp username in the link itself...so its not prompted.

If you have your own server, or are getting it hosted, use SCP to transfer your files securely to a shelled user. Setup apache/httpd and host your images, its much easier than ftp

Best of luck.