I am not a carpenter, I am building a blackjack table and am designing the
base and have a question or two about miter(mitre?) cuts.
The frame for the base is going to be built out of 2x4, possible 2x6's.
I need this to be 38 inches in height, but I want the support legs angle in
at about a 30 degree angle. How do I determine the length of the stock that
I need so that retain the height I need?

Compound cuts are next !

Thanks
Brandon

Well, not sure how to it mathmatically, but what I would do is miter the
bottom of the leg, mark a level line on wall or something 38" high, hold the
mitered leg next to the wall, mark the height.

dave


"Brandon" wrote in message
newsp7gc.7608$S42.1209@lakeread03...
I am not a carpenter, I am building a blackjack table and am designing the
base and have a question or two about miter(mitre?) cuts.
The frame for the base is going to be built out of 2x4, possible 2x6's.
I need this to be 38 inches in height, but I want the support legs angle
in
at about a 30 degree angle. How do I determine the length of the stock
that
I need so that retain the height I need?

Compound cuts are next !

Thanks
Brandon

When you're doing something like this, then a full-size drawing helps a lot.
A sheet of ply or mdf makes a good surface - opposite sides are parallel,
and the corners are at an accurate 90 degress, so this helps in your layout.

Mark out your stock directly from the drawing.

HTH

Frank


"Brandon" wrote in message
newsp7gc.7608$S42.1209@lakeread03...
I am not a carpenter, I am building a blackjack table and am designing the
base and have a question or two about miter(mitre?) cuts.
The frame for the base is going to be built out of 2x4, possible 2x6's.
I need this to be 38 inches in height, but I want the support legs angle
in
at about a 30 degree angle. How do I determine the length of the stock
that
I need so that retain the height I need?

Compound cuts are next !

Thanks
Brandon




---
Outgoing mail is certified Virus Free.
Checked by AVG anti-virus system (http://www.grisoft.com).
Version: 6.0.659 / Virus Database: 423 - Release Date: 15/04/2004

Make 1, 30 degree cut, set the piece on the floor with the 30 degree cut on
the floor, measure up from the floor your desired distance and mark where
the other 30 degree cut needs to be.


"Brandon" wrote in message
newsp7gc.7608$S42.1209@lakeread03...
I am not a carpenter, I am building a blackjack table and am designing the
base and have a question or two about miter(mitre?) cuts.
The frame for the base is going to be built out of 2x4, possible 2x6's.
I need this to be 38 inches in height, but I want the support legs angle
in
at about a 30 degree angle. How do I determine the length of the stock
that
I need so that retain the height I need?

Compound cuts are next !

Thanks
Brandon

In article op7gc.7608$S42.1209@lakeread03,
Brandon wrote:
I am not a carpenter, I am building a blackjack table and am designing the
base and have a question or two about miter(mitre?) cuts.
The frame for the base is going to be built out of 2x4, possible 2x6's.
I need this to be 38 inches in height, but I want the support legs angle in
at about a 30 degree angle. How do I determine the length of the stock that
I need so that retain the height I need?


first the theory, then a practical answer.

Basic trigonometry gets you there. The 'height' of the angled stock is
the 'sine' of the angle off the _horizontal_, and the distance away from
vertical is the 'cosine' of the angle off horizontal. Note: the 'calculator'
on a Windows PC knows those functions. (under 'view', click 'scientific')
You take the sine of the desired angle (60 degrees, in this case), and
_divide_ the desired height, to get the length of the diagonal.


Now, 30 degrees/60 degrees is a 'special' case. the short side of a 30/60
right triangle is exactly half the length of the diagonal. This makes
the other side "square-root-of-three"/2 of the diagonal. And, conversely,
the diagonal is 2/'square-root-of-three' the vertical distance. square-root-
of-three is 1.732, so half that is .866 Thus, to get a 38" height, you need
a piece that is 38/.866 inches long which works out to a hair over 43-7/8
inches.

On Sat, 17 Apr 2004 03:12:52 -0700, "Brandon"
wrote:

I am not a carpenter, I am building a blackjack table and am designing the
base and have a question or two about miter(mitre?) cuts.
The frame for the base is going to be built out of 2x4, possible 2x6's.
I need this to be 38 inches in height, but I want the support legs angle in
at about a 30 degree angle. How do I determine the length of the stock that
I need so that retain the height I need?

Compound cuts are next !

Thanks
Brandon


From your description I can't clearly picture in my sluggish Saturday
morning brain what you need to figure. Some reasonably straightforward
trigonometry will be able to give you the answer, however.

Since I don't have a real helpful suggestion, my point would be that
even after doing the trigonometry, real world conditions are never
what they were supposed to be on paper. Your stock will be somewhat
thinner, thicker, or wider than you intended, your floor isn't flat
and level, your saw's miter gauge is only accurate to 1/2 degree,
etc., etc. (the last being a very important example, since over 38
inches an error of 1/2 degree in that 30 degree angle could become a
7/16" error).

I'd suggest leaving the support legs long until they've been fitted or
even attached, and then marking the length and cutting them off.
That's pretty much foolproof, and singularly deals with all the little
inevitable and cumulative errors that occur when working with many
angled joints.

I don't subscribe to the excuse that wood is an amorphous substance
whose dimensions change over time. Although it is true, it is not an
excuse for poor tolerances and sloppy joints. Just the same, no matter
how small your tolerances are when working wood, especially with
angled joints, your materials and equipment will throw you curves, so
it's much better to expect some of them and work with them than it is
to fight them every step of the way...

John

John Paquay


"Building Your Own Kitchen Cabinets"
http://home.insightbb.com/~jpaquay/shop.html
------------------------------------------------------------------
With Glory and Passion No Longer in Fashion
The Hero Breaks His Blade. -- Kansas, The Pinnacle, 1975
------------------------------------------------------------------

"Brandon" wrote in message news:op7gc.7608$S42.1209@lakeread03...
I am not a carpenter, I am building a blackjack table and am designing the
base and have a question or two about miter(mitre?) cuts.
The frame for the base is going to be built out of 2x4, possible 2x6's.
I need this to be 38 inches in height, but I want the support legs angle in
at about a 30 degree angle. How do I determine the length of the stock that
I need so that retain the height I need?

Compound cuts are next !

Thanks
Brandon

SOH CAH TOA

Check out a high school trigonometry or geometry book. Heres a basic
site, but DAGS on trigonometry, sin, cos, tan.

http://www.mathsrevision.net/gcse/sin_cos_tan.php

Jay

Brandon wrote:
I am not a carpenter, I am building a blackjack table and am
designing the base and have a question or two about
miter(mitre?) cuts. The frame for the base is going to be
built out of 2x4, possible 2x6's. I need this to be 38 inches
in height, but I want the support legs angle in at about a 30
degree angle. How do I determine the length of the stock that
I need so that retain the height I need?

Compound cuts are next !

I put together a web page with some of the trig formulas I've
used most at http://www.iedu.com/DeSoto/trig.html

--
Morris Dovey
DeSoto, Iowa USA

On Sat, 17 Apr 2004 03:12:52 -0700, "Brandon"
wrote:

I am not a carpenter, I am building a blackjack table and am designing the
base and have a question or two about miter(mitre?) cuts.
The frame for the base is going to be built out of 2x4, possible 2x6's.
I need this to be 38 inches in height, but I want the support legs angle in
at about a 30 degree angle. How do I determine the length of the stock that
I need so that retain the height I need?

Cut the miter first, then measure down [along, if on a worktable by
now] and cut the base to length.

Compound cuts are next !

I could post a spreadsheet to do any angles/ No of sides, but can't
send binaries here. Where would it go if not here? If needed, state
if you want in degrees, or units of 5 degrees.... [Can do it to
thousandths of a degree for the whackos who post about their refined
saw cuts, but won't.]

I might just send the formula if someone wants to use just that then
DIY.

Dan.

On Sat, 17 Apr 2004 03:12:52 -0700, "Brandon"
wrote:

I am not a carpenter, I am building a blackjack table and am designing the
base and have a question or two about miter(mitre?) cuts.
The frame for the base is going to be built out of 2x4, possible 2x6's.
I need this to be 38 inches in height, but I want the support legs angle in
at about a 30 degree angle. How do I determine the length of the stock that
I need so that retain the height I need?


OK, I have a better picture of what you mean, but don't know how this
will look at your end:

Thank you all very much for the information. I have an HP scientific
calculator that i am able to figuire this out on, but in practicality, just
doing it with a full size drawing is going to work best.
Thanks again !
Brandon

"Brandon" wrote in message
newsp7gc.7608$S42.1209@lakeread03...
I am not a carpenter, I am building a blackjack table and am designing the
base and have a question or two about miter(mitre?) cuts.
The frame for the base is going to be built out of 2x4, possible 2x6's.
I need this to be 38 inches in height, but I want the support legs angle
in
at about a 30 degree angle. How do I determine the length of the stock
that
I need so that retain the height I need?

Compound cuts are next !

Thanks
Brandon

This depends a bit on how you are measuring your 30 degrees. If it is 30
degrees off the horizontal, then the length of the leg sides will be 38
inches times the tangent of 30 degrees. This would get you to the floor.

Unfortunately, I've decided that I can see about a dozen different ways you
might be doing this, and each has a different answer. What you might
consider doing is to draw a cross sectional picture of the table, either
full size or to scale, showing those legs. Make sure you include the
dimensions of the legs, because the answers also vary if you are using
2x4's, 2x6's, or dowels (which would look pretty spiffy), mainly because of
the width of the wood. After you have a good cross section made up, make
the legs to meet the drawing. Note that you also have to consider how the
leg is attached to the frame and the base.

Michael

"Brandon" wrote in message
news:hsfgc.8059$S42.7550@lakeread03...
Thank you all very much for the information. I have an HP scientific
calculator that i am able to figuire this out on, but in practicality,
just
doing it with a full size drawing is going to work best.
Thanks again !
Brandon

"Brandon" wrote in message
newsp7gc.7608$S42.1209@lakeread03...
I am not a carpenter, I am building a blackjack table and am designing
the
base and have a question or two about miter(mitre?) cuts.
The frame for the base is going to be built out of 2x4, possible 2x6's.
I need this to be 38 inches in height, but I want the support legs angle
in
at about a 30 degree angle. How do I determine the length of the stock
that
I need so that retain the height I need?

Compound cuts are next !

Thanks
Brandon

In rec.woodworking
"Brandon" wrote:

Thank you all very much for the information. I have an HP scientific
calculator that i am able to figuire this out on, but in practicality, just
doing it with a full size drawing is going to work best.

Practicality? Do you have access to a large plotter or are you going to
tape a bunch of pages together? You want 30 degree slant legs to rise 38"?
This is 8th grade stuff.

Oliver had a heap of apples.

Sin = O/H
Cos = A/H
Tan = O/A

To determine the length of the hypotenuse with a side of 38" and an angle
of 60 degrees(which is what I think you mean, not 30, you'd can use sin:

38 = c * sin 60

therefore
c = 38/sin 60

c = 43.878

Cut your boards 43-7/8" to the long point of the miter cut and you're
perfect, if I understand your design.

If you really do mean 30 degrees, the boards are going to be 76" long.

Bruce,

76 inches is quite along leg. I'd be concerned of it bending. The
other thing to think about is that these legs might go from a base (of
undetermined height and attachment), to some location presumably under the
38 inches. Oliver and his apples are good once the problem is completely
described.

Michael
Measure twice cut once. Measure again, cut again. Measure a third time,
discover that the first cut was right.

"Bruce" wrote in message
.. .
In rec.woodworking
"Brandon" wrote:

Thank you all very much for the information. I have an HP scientific
calculator that i am able to figuire this out on, but in practicality,
just
doing it with a full size drawing is going to work best.

Practicality? Do you have access to a large plotter or are you going to
tape a bunch of pages together? You want 30 degree slant legs to rise
38"?
This is 8th grade stuff.

Oliver had a heap of apples.

Sin = O/H
Cos = A/H
Tan = O/A

To determine the length of the hypotenuse with a side of 38" and an angle
of 60 degrees(which is what I think you mean, not 30, you'd can use sin:

38 = c * sin 60

therefore
c = 38/sin 60

c = 43.878

Cut your boards 43-7/8" to the long point of the miter cut and you're
perfect, if I understand your design.

If you really do mean 30 degrees, the boards are going to be 76" long.

Here's a nice site for woodworking related downloads, including one Badger
Pond used to list for cutting compound miters.

http://www.woodcentral.com/bparticle...html#downloads

On Sat, 17 Apr 2004 21:01:04 GMT, (Bruce) wrote:

If you really do mean 30 degrees, the boards are going to be 76" long.

Not likely!

Another approach in this particular case:

A right triangle with 30,60 other angles has sides in the ratio
1:2:sqrt(3). The 30 deg has to be at the top or the bottom. He
likely means the top, or the table is squat.

In that case, the other non-right angle is 60, at the bottom.

With a vertical side of 38", the slope is twice that divided by
sqrt(3), or 43.88" [ ~43 7/8].

Dan.

In article ,
Danny Boy wrote:
On Sat, 17 Apr 2004 21:01:04 GMT, (Bruce) wrote:

If you really do mean 30 degrees, the boards are going to be 76" long.

Not likely!

Yes, TRUE.

*IF* the leg angle is 30 degrees off the _horizontal_.


Another approach in this particular case:

A right triangle with 30,60 other angles has sides in the ratio
1:2:sqrt(3). The 30 deg has to be at the top or the bottom. He
likely means the top, or the table is squat.

In that case, the other non-right angle is 60, at the bottom.

With a vertical side of 38", the slope is twice that divided by
sqrt(3), or 43.88" [ ~43 7/8].

And if the tables _is_ squat, to use your terms, then the leg length
*is* 6' 4".

Dan.

In rec.woodworking
Danny Boy wrote:

On Sat, 17 Apr 2004 21:01:04 GMT, (Bruce) wrote:

If you really do mean 30 degrees, the boards are going to be 76" long.

Not likely!

As I tried to imply in my message.

Another approach in this particular case:
With a vertical side of 38", the slope is twice that divided by
sqrt(3), or 43.88" [ ~43 7/8].

Strangely enough, the exact same number I came up with. Did you have a
point? The rules about 30-60 triangles are nothing but memorized
derivitives of the sin rules.

On Sun, 18 Apr 2004 03:33:04 GMT, (Bruce) wrote:

Hi,

Strangely enough, the exact same number I came up with. Did you have a
point? The rules about 30-60 triangles are nothing but memorized
derivitives of the sin rules.

I was just giving an alternate approach, for God's sake. The "rules"
are what makes it work, and they came from an 'understanding' of the
working principles, ratio of sides. It is a matter of understanding,
not of memorising. You learned to do fractions by looking at the
simplest cases first (1/2, 1/3, 1/4, ....) then you learned the rules
you could apply to any other numbers. It's no different here.

In any case, you can figure it out for yourself. I mentioned the
30,60,90 only because it applied in this particular case, as I already
said. That made an easier approach possible. If you stick to just
one method, it's like having just one hammer in your toolbox. You can
ask about each problem as you meet it, or learn the basic principles
and apply them anytime. Your choice.

Dan.

Bruce wrote:

In rec.woodworking
"Brandon" wrote:

Thank you all very much for the information. I have an HP scientific
calculator that i am able to figuire this out on, but in practicality,
just doing it with a full size drawing is going to work best.

Practicality? Do you have access to a large plotter or are you going to
tape a bunch of pages together?

Perhaps he has access to a pencil?

You want 30 degree slant legs to rise
38"? This is 8th grade stuff.

Oliver had a heap of apples.

Sin = O/H
Cos = A/H
Tan = O/A

To determine the length of the hypotenuse with a side of 38" and an angle
of 60 degrees(which is what I think you mean, not 30, you'd can use sin:

38 = c * sin 60

therefore
c = 38/sin 60

c = 43.878

Cut your boards 43-7/8" to the long point of the miter cut and you're
perfect, if I understand your design.

If you really do mean 30 degrees, the boards are going to be 76" long.

--
--John
Reply to jclarke at ae tee tee global dot net
(was jclarke at eye bee em dot net)

In rec.woodworking
"J. Clarke" wrote:

Bruce wrote:

In rec.woodworking
"Brandon" wrote:

Thank you all very much for the information. I have an HP scientific
calculator that i am able to figuire this out on, but in practicality,
just doing it with a full size drawing is going to work best.

Practicality? Do you have access to a large plotter or are you going to
tape a bunch of pages together?

Perhaps he has access to a pencil?

LOL! Yeah, maybe so.