Trig, as in trigonometry

[8thsinner]

Can anyone provide the equation of calculating an edge angle using only the dimensions of the triangle without knowing any angles, I want to know this for my new TTi system of sharpening. It's been over 13 years since leaving school so this area of my mathematics is quite minimal...I am fairly sure at least of the soh, cah, toa acronym holds true, but be damned if I can remember how to use it espceially using a mobile phone calculator...which operaters differently to the logic operated system (which had it's own name at the time) I was used to using at school.

I have searched google a little, but as I am using my mobile phone as my modem data charges can be quite excessive, and I cannot afford them.

My triangle sizes are thus...roughly measured
Adjacent = 132mm
Opposite = 40mm
Hypotenuse = 148mm

If you can be bothered to calculate this for me I would appreciate it...but also leave me the formulae please.

Thanks

 

[steveme]

I'm a bit rusty on this stuff myself, but I think it's Sin A = opp / hyp (from soh-cah-toa)

so :

40 / 148 = 0.270270

arcsin 0.270270 = 15.68 degrees

which on a triangle that has long adjacent and hypotenuse sides and a relatively narrow opposite seems about right.

does that work?

 

[8thsinner]

thats the thing, I really don't know, anything between 13 and 20 or so would sound right to me...
I would have estimated the angle to be about 20 degrees , but it's real hard to tell with things like this.

anyone else wanna jump in?

 

[Aussiepom]

As already mentioned, you can work it out using any 1 of the 3 combinations of SOH, CAH, TOA, and they should all come out to the same answer.

The figures that you've provided give answers of 15.68, 26.888 and 16.858 respectively. This means one of 2 things: Either the dimensions you've provided are incorrect, or the triangle is not a 'right angled' triangle. Trig only works on 'right angled' triangles.

Of course, it's been even longer since I was at school, so if anyone else wants to chime in, that's fine with me. (Also, I'm doing this in a mad rush at the moment, so I could easily have made a mistake.)

 

[8thsinner]

I will attempt to make more accurate measurements to try and rectify this error possibility, but even these are judgment based using literally a ruler and rough placement of what I think should be the point of measurement. A/O placement...

This is a lot more complicated than I remember it being....

Anyone else...?

 

[Toddy]

Do you know how long it is since I did this stuff ??

That said, from the figures you give I reckon sinA, so sin = opposite over hypotoneuse. And Steve's done the work 15.68 degrees

cheers,
Toddy

 

[FGYT]

Just measure the Opposite and adjacent

then its Tan = Opp/Adj

we always used

The = Old/ Arab
Sits = On/His
Camel =And/Howls

because our old Maths teacher always wor a Arab robe etc to school fancy dress stuff

however i just draw it on Auto CAD these days and measure it

your measurements 40 & 132 give
16.86 deg but this is a Hyp of 137.93

 

[VirusKiller]

Trig only works on 'right angled' triangles.

Quite correct. However, there is an extension to arbitary triangles:

http://en.wikipedia.org/wiki/Law_of_cosines

Images taken from http://en.wikipedia.org/wiki/Trigonometry.

 

[Grooveski]

15.07°

 

[Native Justice]

I'm sure this will help.

 

[Landy_Dom]

I'm sure this will help.

SPOT ON MATE

i was thinking of elaborating on the t/2 issue but you've explained it beautifully. Should be a useful tool for us folks on here

Dom.

 

[K.NYPH]

I do love to see people talking about something we were supposed to have hated at school,sorry I loved it.
Now for your next project you can work out the edge angle on a convex blade using euclidian trigonometry (angles on curved surfaces)
Ok I'm a smart ****.

 

[Toddy]

Did we ever reach a definitive answer though ?

cheers,
Toddy

 

[8thsinner]

I think you mean euclidean trigonometry...smart bottom...
lol
ANd yeah I think we did toddy, I am at least happy enough with fifteen degrees...it's field work, I don't need decimals

 

[8thsinner]

Actually, you got me thinking.

I would be very interested despite the fact of not understanding it, of seeing a comprhensive study of curvature of blades in reference to efficiency of cutting and strength of edge and edge holding properties.

ANd too, considering I want to start making knives, and I will one of these days, it's a good thing to learn to calculate, the regular trig and hyperbolic trig...

 

[K.NYPH]

I said I was good at maths,english was a disaster zone,but thank you for the correction.

 

[K.NYPH]

As for the difinative answer to the original question:
1. Make life simple halve the blade thickness.
2. Use the sin rule:- A/sin a = B/sin b = C/sin c
in conjunction with the Pythagoras Theorem :-
C squared = A squared + B squared , where C is the hypotenuse
3. You know A & B Which are lengths.
4. Use Pythagoras to find C
5. You know one angle which is 90 degrees (sin 90 is 1)
6. Therefore the sine rule is modified to:-
C/sin90 = A/sin a
which becomes
C/1 = A/sin a
Now if you transpose the equation
sin a(-1) = A / C

Q.E.D

 

[charadeur]

http://www.wolframalpha.com/
Check out Wolfram Alpha. Math teachers hate that site.

 

[wildman695]

who really gives a damn what the angle is as long as it cuts, hee hee

 

[Air Pirate]

Seems to me that the angle of the root flange on your canuder valve is all cattywumpus.

You should probably get that looked at.