Essentially, I need to know the angle formed by the intersection of two planes. Both with 6° elevation and with the planes 90° to each other.

Does anyone have the solution and possibly the derivation of the formula?

I need to drill the holes tomorrow if possible.

Any help would be appreciated.

Preston

It will be interesting to see the decoding of this question.

If you elevate adjacent sides of a solid 'slab' it will be on it's corner.

??

For now I'm guessing 6 degrees.?

--

Mark

N.E. Ohio

Mark

N.E. Ohio

Click to see the full signature.

Actually, if anyone has a machinists handbook the answer is in there.
The resulting angle from the first two angles has to be calculated (it's
not 6 degrees). I have my copy at work, or else I could give yo the
answer right now. Help anyone??????

Mark wrote:

Mark wrote:

I may have to wait a day to drill until I get the answer. Perhaps you can
post it tomorrow.

Thanks,

Preston

message

I

Thanks,

Preston

message

I

I'll lay odds that for an angle this small, it's within half a degree of 6
degrees.

todd

message

I

corner.

todd

message

I

corner.

Having done the math, I'll take you up on that.

What odds are you giving, I want $100 worth? <grin>

The odds are 1:1000. If you send me the $100, I'll get your $10 out right away.

todd

Yes it will be on the corner. The angle from the corner touching the table
to the opposite corner is what I need and it isn't 6°

Preston

side.

I

the

Preston

side.

I

the

I'm not sure if I understand what you're trying to do, but here goes:

Assuming the 'original' methodology is: elevate one side (call it AB)of the slab, so that a line from it to the other side (call it CD) is 6 degrees above horizontal. then elevate side BC so that a line from

And what you want is the angle above horizontal of the line BD. B is the high corner, and D is still at the original horizontal.

Taking things piecemeal, corner "A" is above the original horizontal plane by sin(6 degrees)

(sin(6 deg)

if the slab is a

theta = arcsin( sqrt(2)*sin(6 deg) )

which is approximately: 8.5009361422462 degrees

Note: when dealing with small angles like this, one

I agree with Horatio's answer, if you elevate one edge (AB) and then elevate
(BC) you essentially raise the height of the top corner twice and you get
8.5 degrees elevation along the diagonal. However, I interpreted the
question slightly differently. If you raise AB by 6 degrees, then the high
corner is at some height X. Now put the slab flat again, elevate BC by 6
degrees, and the height of the top corner is again at the same height X. In
order for the high point to be at the same position for both cases, the slab
must be a square. Not surprisingly, with my interpretation of the problem,
the elevation along the diagonal is 1/2 of Horatio's answer. (4.24
degrees.)

wrote:

side.

the

AB)

AB)

percent

wrote:

side.

the

AB)

AB)

percent

At the risk of flogging a dead horse a few days late, here's an easier way
to get the 4.24 degree answer:

Consider a square labeled like a baseball diamond. We want to raise the 2nd base corner while leaving home plate alone. From the original question, I assume that we want to tilt the square such that the 1st, 2nd and 3rd base corners are all raised to the same height as each other. If the distance from home plate to 1st base is L, then the distance from home plate to 2nd base is L***sqrt(2). Both 1st and 3rd base will be elevated by L***sin(6
degrees), while 2nd base will be elevated by L***sqrt(2)***sin(theta). We want
to know what value of theta will make the heights the same (L***sin(6 degrees)
= L***sqrt(2)*sin(theta)). This occurs for theta = arcsin[ sin(6 degrees) /
sqrt(2) ] = 4.2388 degrees.

BadgerDog

wrote:

I

formula?

side

disregard

the

Consider a square labeled like a baseball diamond. We want to raise the 2nd base corner while leaving home plate alone. From the original question, I assume that we want to tilt the square such that the 1st, 2nd and 3rd base corners are all raised to the same height as each other. If the distance from home plate to 1st base is L, then the distance from home plate to 2nd base is L

BadgerDog

wrote:

I

formula?

side

disregard

the

BadgerDog wrote:

Easier... Heh.

Being a mathematical retard sucks. What is that, trigonometry or something?

Easier... Heh.

Being a mathematical retard sucks. What is that, trigonometry or something?

--

Michael McIntyre ---- Silvan < snipped-for-privacy@users.sourceforge.net>

Linux fanatic, and certified Geek; registered Linux user #243621

Michael McIntyre ---- Silvan < snipped-for-privacy@users.sourceforge.net>

Linux fanatic, and certified Geek; registered Linux user #243621

Click to see the full signature.

A clarification:

Re-reading my response I don't think I described the process correctly and what I was trying to say would most likely be mis-interpreted due to its poor explanation.

I was primarily trying to show an easier way to do the calculation that Horatio presented in the piece meal approach. There is also the problem of more than one interpretation of the problem. So here is a brief description of the easier calculation for either interpretation:

1) Horati's interpretation: the "top corner" is effectively raised twice as high as the corners that are defined by the 6 degree incline. If "L" is the length of the side of the square, this interpretation means that L***sqrt(2)***sin(theta) = 2***L***sin(6 degrees), where theta is the angle if
inclination for the top corner and the length of the diagonal of the square
is sqrt(2)***L. This gives theta = arcsin[ 2 *** sin(6 degrees) / sqrt(2)] 8.5009 degrees.

2) Michael's interpretation (which I described poorly) was that we wanted to raise the top corner the same amount as the other corners when they are raised to an inclination of 6 degrees. In this case, L***sqrt(2)***sin(theta) L*sin(6 degrees), which occurs for theta = arcsin[ sin(6 degrees) / sqrt(2)]
= 4.2388 degrees (as I described in the previous post).

I am afraid that my description in the original post implies that 1st, 2nd and 3rd base are raised to the same height at the same time; this would just raise the square, not tilt it.

BadgerDog

wrote:

reasons,

elevate

two

other.

side

Re-reading my response I don't think I described the process correctly and what I was trying to say would most likely be mis-interpreted due to its poor explanation.

I was primarily trying to show an easier way to do the calculation that Horatio presented in the piece meal approach. There is also the problem of more than one interpretation of the problem. So here is a brief description of the easier calculation for either interpretation:

1) Horati's interpretation: the "top corner" is effectively raised twice as high as the corners that are defined by the 6 degree incline. If "L" is the length of the side of the square, this interpretation means that L

2) Michael's interpretation (which I described poorly) was that we wanted to raise the top corner the same amount as the other corners when they are raised to an inclination of 6 degrees. In this case, L

I am afraid that my description in the original post implies that 1st, 2nd and 3rd base are raised to the same height at the same time; this would just raise the square, not tilt it.

BadgerDog

wrote:

reasons,

elevate

two

other.

side

BadgerDog wrote:

I didn't understand the question, once I did a practical solution was easy.

I didn't understand the question, once I did a practical solution was easy.

--

Mark

N.E. Ohio

Mark

N.E. Ohio

Click to see the full signature.

Preston Andreas wrote:

Assuming you've got a square slab with corners 1,2,3 & 4 with corner 1 remaining on the original plane and you tilt corner 2 up 6 degrees then holding corners 1 & 2 at this position and elevate corner 3, diagonally accrossed from corner 1, up 6 degrees the resulting angle between corners 1 & 3, relative to the original plane is 8.5 degrees or close enough to 8.5 degrees

Anyone want to confirm my results so I don't feel too bad if Preston uses 8.5 degrees and I'm wrong. Had to remember how to convert degrees to radians and radians back to degrees.

charlie b

Assuming you've got a square slab with corners 1,2,3 & 4 with corner 1 remaining on the original plane and you tilt corner 2 up 6 degrees then holding corners 1 & 2 at this position and elevate corner 3, diagonally accrossed from corner 1, up 6 degrees the resulting angle between corners 1 & 3, relative to the original plane is 8.5 degrees or close enough to 8.5 degrees

Anyone want to confirm my results so I don't feel too bad if Preston uses 8.5 degrees and I'm wrong. Had to remember how to convert degrees to radians and radians back to degrees.

charlie b

Thanks for the help. I made the jig with about 8.5° elevation. I will
drill tomorrow. Hope it works.

Preston

Preston

I hope you have already figured this out, but here it goes. You need a center line for your slab. Mark your hole location and draw a line through the center of that hole perpendicular the slab center line. Since your angles are both six degrees your sight line will be a line that extends from the hole center to the slab center line at 45 degrees to the hole center line perpendicular. Here is the tricky part. You dont elevate a corner. You elevate the far side of the sight line. To get 6 degrees on both planes tilt your drill pres table to 8.5 degrees. Keep the sight line perpendicular to the table tilt and you will be fine. I got this information out of the back of Drew Langsner's book The Chair Makers Workshop. It has tables for all the possible rake and splay angles and their sight lines. It also has a better explanation of the process than I have given here.

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