Hi gang,

I need to know how to find a particular angle without any special tools. I didn’t write down some of the angles on a light when I made them but now I need to know what they are.

How do I find what that angle is based on the measurements?

—

In Him (Jesus Christ) was life; and the life was the light of men. And the light shineth in darkness; and the darkness comprehended it not.

http://asflashlights.com/ High powered super lights and and parts!

http://www.1728.org/quadtrap.htm

-Clark

tan x = ((0.885-0.785)/2)/(0.090)

x = arctan(((0.885-0.785)/2)/(0.090)) = 29°

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Hi,

I’m not sure which angles you mean exactly, but here’s a crappy pic I made from yours:

If you are looking for ‘alpha’ then you need to calculate tangent of this angle which is 0.050/0.090=0,55555(5)

If you have this value, you go to online calc and check, what angle value corresponds to this result, and for tg=0,55555(5) the angle is 29,0546°

Does 29 degrees sound right?

My current and or voltage measurements are only relevent to anything that I measure.

Budget light hobby proudly sponsored by my Mastercard and unknowingly paid for by a hard working wife.

djozz said "it came with chinese lettering that is chinese to me".

old4570 said "I'm not an expert , so don't suffer from any such technical restrictions".

Old-Lumens. Highly admired and cherished member of Budget Light Forum. 11.5.2011 - 20.12.16. RIP.

The angle of the sides of the trapezoid with reference to the “body” should be 146.25 or 33.75 for the angle coolperl illustrated. Right?

edit: hurrdurr, derp, etc.; .090 corresponds to leg height not hypotenuse length….durr I feel so stoopid-Clark

Both ,X3 and coolperl are correct ! 29 degrees !

Adrian

Nope. That would be the case, if the 0,090 value was the lenght of the diagonal part. On the picture this value is lenght of vertical projection of diagonal part.

Or in another words, your values are calculated from the sine function, but should be from tan (or tg) function, because only dimensions for tangent function are provided.

This is correct. The angle is 29 degrees.

My current and or voltage measurements are only relevent to anything that I measure.

Budget light hobby proudly sponsored by my Mastercard and unknowingly paid for by a hard working wife.

djozz said "it came with chinese lettering that is chinese to me".

old4570 said "I'm not an expert , so don't suffer from any such technical restrictions".

Old-Lumens. Highly admired and cherished member of Budget Light Forum. 11.5.2011 - 20.12.16. RIP.

29 degrees would make sense, since that is where my compound sits by default most of the time. I think this is exactly what I needed.

So to be sure I understand. Half the distance of the change in diameter divide by the distance traveled. Then plug that number into the inverse tangent calculator and that is my angle?

So if all other things were the same but the head diameter was 1.250” I would go… 0.2325/0.090=2.5833333 For an angle of 68.83?

Also why in the equation does the decimal turn to a comma? 0.050/0.090=0,55555(5)

Thank you

In Him (Jesus Christ) was life; and the life was the light of men. And the light shineth in darkness; and the darkness comprehended it not.

http://asflashlights.com/ High powered super lights and and parts!

USA: 1,000 = thousand

Other countries (including your calculator): 1,000 = one point zero zero zero

So the “.” is also reversed.

Correct!

(P.S. make sure your calculator is set in “deg” instead of “rad” or “grad”)

Sorry, that’s a typo. Ignore it.

LOL… Shows how much I know about trig. you could have put %(*()#@)#!!)) or kldswiqeorpugjriwqfpeiofj and I would have assumed it meant something.

In Him (Jesus Christ) was life; and the life was the light of men. And the light shineth in darkness; and the darkness comprehended it not.

http://asflashlights.com/ High powered super lights and and parts!