Calculation Help!

[claym]

X=Y=204xCos(30)=177 cm
L1=L3=204xSin(30)=102 cm

L2 fås genom 204-L2=102 cm
L2=sqrt(102^2+102^2)=144 cm
Man kan ta 102/cos(45) också

Edit: lite seg på att skriva

Hmm... does this really add up?

The formula states cos(a) = adjacent/hypotenuse
The adjacent is 204, the hypotenuse is x

https://sv.m.wikipedia.org/wiki/Cosinus

 

[richardtenggren]

Well... does this really add up?

The formula says cos(a) = adjacent/hypotenuse
The adjacent is surely 204, the hypotenuse x

You're right, a bit sleepy this morning
Mixed it up, it should be as you wrote:
204/cos(30)=236

 

[richardtenggren]

X=Y=204xCos(30)=177 cm
L1=L3=204xSin(30)=102 cm

L2 obtained through 204-L2=102 cm
L2=sqrt(102^2+102^2)=144 cm
You can also take 102/cos(45)

Edit: a bit slow to write

I was newly awake The above fell on its own merit...
X=Y=204/Cos(30)=236 cm
L1=L3=204*Tan(30)=118cm

L2 obtained through 204-L3=86 cm
L2=sqrt(86^2+86^2)=122cm
You can also take 86/cos(45)

 

[camaro1969]

Hmm
But if I run a 30 degree angle up to the left..

Then the angles at the very ends of the steps become uneven..
60 on one and 75 on the other..?

I wanted these to be at 22.5 degrees...:/

 

[claym]

I was just awake The above fell on its own reasonableness...
X=Y=204/Cos(30)=236 cm
L1=L3=204*Tan(30)=118 cm

L2 is obtained through 204-L3=86 cm
L2=sqrt(86^2+86^2)=122 cm
You can also take 86/cos(45)

Calculating trig in the morning is a bit tough, even I had to think about it a couple of times .

Off topic: I had this discussion with a person 5 years ago; he also rearranged the calculations. To this day, he claims that he calculated it correctly. Today he works as a professor of mechanics at a university....

 

[claym]

Hmm But if I run 30 degrees angle ippe to the left.. Then the angles at the outermost step will be uneven.. 60 on one and 75 on the other..? I would probably want these to be at 22.5 degrees...:/

But if you want to lock "60" and "75" to 22.5, you will get more "lines" going out. In that case, you will have the top angle change from 30 to: 180-22.5-90=67.5

 

[camaro1969]

But if you want to lock "60" and "75" to 22.5, you will get more "lines" going out

What do you mean?

 

[claym]

How do you mean?

Sorry not more. You get a top angle which is 67.5

 

[camaro1969]

I don't understand anything anymore.

 

[Robert-san01]

Going to build today..
Cad available tomorrow

You need to be a bit more careful when drawing.

 

[richardtenggren]

That thing with sin, cos, and tan often confuses things, especially now when they're rarely used.
You should always double-check...

Someone who taught me about machining drew the rollers' rotation direction incorrectly both in their thesis and in our lectures many years later.

 

[claym]

That thing with sin, cos, and tan can easily confuse, especially now when you use it so rarely. You should always double-check...

Someone who taught me in processing drew the direction of the rollers' rotation incorrectly both in his thesis and in our lectures many years later.

 

[claym]

I don't understand anything anymore

"The 'problem' arises because you want to lock one angle to 22.5 and an angle is already locked to 90 degrees, the result is that the last angle becomes 67.5 (i.e., the top angle). Since I assume you want the construction 90 degrees between the 204 sides, you won't be able to symmetrically fit this if you lock at 22.5."

 

[camaro1969]

It sounds really...

So you have to compromise on symmetry, and the "corner" becomes larger than the "ends"?

What are the measurements if I want to lock at 22.5?

 

[claym]

That sounds right..

So you have to compromise on the symmetry, and the "corner" becomes larger than the "ends"?

What are the measurements if I want to lock at 22.5?

One of the top angles becomes 67.5 and the other becomes 22.5.

Cos(67.5) = 204/y
Y = 204/cos(67.5)

Cos(22.5) = 204/x
X = 204/cos(22.5)

But perhaps I misunderstood, I realize? You don't want one at 22.5 and the other at 67.5, do you?