Math Formula Required

[Kiwi]

I need to calculate the length of the side marked (?) of an oblique triangle.
All the formula I have don't match the known facts.
This is all I know, can the marked side be calculated?
Thankyou.

[Ken_Shea]

Kiwi,
Because I do not understand much of algebraic formulas, I have calculated similar needs right in my cad, this one is very simple.

This is in imperial so you will need to convert to Metric.

[NC Cams]

The angle inside the circle is 13.927123 deg

The length of the segment you'r looking for is 19.924348

Done via DesignView version DV50-10gb (English version), a superb paramertic CAD sketchpad program that I wouldn't trade for the world

RE-EDIT Great program providing you connect the dots properly - see later post END RE-EDIT

EDIT

To calculate, first drop a perpendicular line from the 6" radial line that is not colinear with the coliniear 20" and 6" lines. Extend it from the OD intersection up to the 20" and 6" colinear lines.

Now that you have a 90 deg angle, you should be able to figure out the lenght of the perpendicular drop.

Having this, you can figure the short length of the segment from the circle OD to the perpendicular drop/rise. Call this 6-x where x is the itsy bitsy length from circle OD to the construction line.

With the 5 deg angle and the length of the perpendicular and the true length of the '20+x' segment you should be able to do the calcs needed to find ?

It was easier and faster to use DesignView

END EDIT

[Ken_Shea]

Hmmmm....
What I get is 20.2692 for a 2D length, actual line length 20.345

Angle inside circle 17.1893

[Caprirs]

Using math only, you could find the equation for the circle:

(x-26)**2 + (y-0)**2 = 6**2

and the equation for the line:

mx + 0 = y where "m" is the slope of a -5 degree line

Substitute the line equation (already solved for y) into the circle equation. You will get two solutions since the line will intersect the circle twice.

[NC Cams]

MAJOR OOPS, Sorry.

Ken Shea is right on at 20.345. (20.345433 to 6 decimals)

The internal angle is 17.190 (17.189734 to 6 decimals)

When I resized the drawing I didn't catch that the CAD program picked up a different arc contact point which it does occasionally on a "resized" sketch.

This proves that you can't always trust what the CAD says - you MUST do a reality check which I sadly failed to do in this instance.

First DUH moment this week - probably won't be the last.

[Geof]

Originally Posted by Caprirs Using math only, you could find the equation for the circle: (x-26)**2 + (y-0)**2 = 6**2 and the equation for the line: mx + 0 = y where "m" is the slope of a -5 degree line Substitute the line equation (already solved for y) into the circle equation. You will get two solutions since the line will intersect the circle twice.

According to the time stamp on the post you are telling someone to solve a quadratic equation before 7 o'clock in the morning!!! Have a heart

[lakeside]

Originally Posted by Geof According to the time stamp on the post you are telling someone to solve a quadratic equation before 7 o'clock in the morning!!! Have a heart

Geof you maybe be the only one who would know the math even before 7:00AM

[fpworks]

Another math dork method use the Law of Cosines:

6^2 = 26^2 + ?^2 - 2*26*?*cos(5 deg)

solve for '?' = 20.3454 AND 31.4567 (with my HP, because I'm lazy)

Common sense (geometry of a triangle) throws out 31.4567

[lakeside]

20.3454 how that # sound

[Kiwi]

Thanks to all who replied.
I can do it on my Cad program OK, I needed the formula to build in a VB program which calculates the cusp height on a curved surface.
Caprirs and fpworks have supplied what I need.
I'll now go and build into my program and see if I get the correct answer.
Also study NC Cams method.

[Kiwi]

NC Cams
Quote Now that you have a 90 deg angle, you should be able to figure out the lenght of the perpendicular drop.>

If I'm reading you right, I don't think I have enough info to calculate the perpendicular line? My attempt to understand attached.