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**Morten Fredriksen****Member**- Registered: 2016-11-30
- Posts: 9

Hello!

First of all:

You have to exuse my english, due to bad skills..but I`m trying.

I see that you guys are into math, for sure, and I have a question for you.

We (Budgetlightforum) are trying to build a super-thrower of a LED flashlight, but in a reasonable handheld size (Tread: www.budgetlightforum.com/node/50303 ).

And we are gathering info about reflectors, and the shape of it..

And here is my question: Which form will be the best shape for this reflector??

The inner diameter, on the outer end is 120mm.

And some my forum have their opinion of 112mm for the depth of it..will be the best.. (But I don`t know their source for that opinion)

Do you have the mathematic answer for my question?

I appreciate for your time to read this, and I even better appreciate an answer from you, if you are willing to.

Bets regard!

Morten Fredriksen

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi;

Why does post #5 not settle the question? Looks like a great use of EM. I have not checked it because I just want to marvel at such sublime beauty and understanding for awhile. First, I must wipe away the tears of joy.

Of course, he did not use the correct tool for the job but that can be fixed easily.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**Morten Fredriksen****Member**- Registered: 2016-11-30
- Posts: 9

bobbym wrote:

Hi;

Why does post #5 not settle the question? Looks like a great use of EM. I have not checked it because I just want to marvel at such sublime beauty and understanding for awhile. First, I must wipe away the tears of joy.

Of course, he did not use the correct tool for the job but that can be fixed easily.

I knew some of you would be laughing..but this doesn't answer my question.

And why I ask is because of all the back and forth with the design of the reflector, some even had the idea of cutting his flashlight (Trunite TN42, who has the record in throw for the moment. 1350meter) reflector i half, and just up scale it to ours measurements.. lol

And then you have this program mentioned in post #5, but it seem like people have not enough knowledge of this program to get what we want...

Maybe some people in this forum has the right tools, or knowledge of this subject??

Looking forward to get the "correct" answer from one of you!

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi;

I was not laughing. I admire his approach very much.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**Morten Fredriksen****Member**- Registered: 2016-11-30
- Posts: 9

What's your view on post #1768 and 1769 at page 59 then?

*Last edited by Morten Fredriksen (2016-11-30 20:06:48)*

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,000

hi Morten Fredriksen

Welcome to the forum.

It is a property of the parabola that rays starting from the focus will reflect off the mirrored surface in parallel rays.** The reverse is also true: rays coming in towards a parabolic mirror will all go through the focus (hence the name). Reflecting telescopes utilise this property to produce focussed images of astronomical objects.

Remember the problems when the Hubble telescope was first launched. In essence, this was due to an incorrect shape for the mirror. https://en.wikipedia.org/wiki/Hubble_Sp … he_problem

If you had a perfect parabolic reflector and a point light source and were shining in a perfect vacuum, the light would go on for ever! In practice you'll never make a perfect mirror, the light source will not be a single point so not all light will originate from the focus, and finally, light will hit dust particles etc and so get scattered.

For a given specification of outer diameter and depth to the centre of the mirror, a parabola can be determined. Then it's a question of how good is your engineering? As well as worrying about the power of the light source, you need to make it as small as possible. And then take it to a high mountain away from cities in still air just after rain, to minimise the dust.

Sorry, I haven't read all those posts; too much this early in the morning (or any time come to think of it)

And your English is very good!

Bob

** I think I can produce a coordinate geometry proof of this if you want.

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

What's your view on post #1768 and 1769 at page 59 then?

I do not have one. Enderman's work in #1767 which is accurate, is about as far as I can understand.

But I do have an interest in your work. I hike alot and sometimes out of necessity those hikes are in the dark. To a predator, night is the equivalent to dusk for a human. Meaning he can see pretty well. I am totally blind in that environment so a good flashlight is mandatory.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**Morten Fredriksen****Member**- Registered: 2016-11-30
- Posts: 9

bob bundy wrote:

hi Morten Fredriksen

Welcome to the forum.

And your English is very good!

Bob

** I think I can produce a coordinate geometry proof of this if you want.

Thank you for welcoming me!

As for the English: When not in use, we forget..but now the knowledge starring to come out from the blur..

(Haven't used my English since high school..)

But a fast learner, and the ability to "adapt"...it can't be worse..

If you are interested in produce that coordinate geometry, it'll be much appreciated!

But I'll throw in an sub-question for that matter: The work of *Endermans* work is not far away from the best equation for this subject?

Thanks in advance!

Morten

*Last edited by Morten Fredriksen (2016-12-01 01:42:24)*

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,000

hi Morten

F (0,f) is the fixed focus on the y axis. D (x,0) is a moving point on the x axis directly below the general point on the parabola, P (x,y). The x axis has a special name for this parabola; it is called the directrix.

A parabola is defined as the locus of all points, P, such that the distance from P to the directrix is equal to the distance of P from the focus. Thus

differentiating with respect to x:

This is the gradient of a tangent AB to the curve at P.

On my diagram, the parabola is shown as a thick line. AB is the tangent to the curve at P. As PF = PD a circle may be drawn, centre P and radius PF = PD. DP extended cuts the circle again at C.

FD has gradient -f/x and so is at right angles to AB. So FPD is an isosceles triangle, bisected by AB.

So if angle FPD = 2 θ, then FPB = θ and using the angle properties of a circle, FCD = θ as well.

Let EP be at right angles to AB. Then CPE = 90 - θ = FPE . These angles are the angle of incidence and the angle of reflection of a light ray coming from C at right angles to the directrix, reflected at P towards F. Thus parallel rays perpendicular to the directrix reflect off the parabolic surface to go through the focus F.

And, making the rays start at F, a parallel beam of light is produced.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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**Morten Fredriksen****Member**- Registered: 2016-11-30
- Posts: 9

Thanks for your answer Bob!

So in this equation/diagram I see that the measures in our flashlight is a bit of..

If we (as for an easily equation) devide the diameter on the hight, 120mm / 112mm, we get 1,0714285714

But that will not be the answer in your diagram?

If I just take a raw measure right from my computer screen, I got (from diagram) a diameter of 114mm, and a hight from F is 88mm = 114/88 = 1,2954545455

So we are slightly of from this diagram in our reflector...

If we do this then:

120/x=1,2954545455

The x will then be = 92,63161795

Can I copy your post Bob, over to our forum?

Or do you want to join our forum and point out this equation/diagram??

The last suggestion would be MUCH-MUCH appreciated!!

Morten.

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,000

hi Morten,

I don't mind my post(with credits ) being copied. Or you could put a link from your forum to ours.

If I call the origin (0,0) point O and the lowest point on the curve L, then OL = FL (from the definition)

And I'll call the point on the y axis level with P, point Q.

Now it looks like you are saying PQ = 120. But where is 112 measured from and to. Most likely from Q but to F or L or even O ?

To use the equation accurately you'll need to work out f.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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**Morten Fredriksen****Member**- Registered: 2016-11-30
- Posts: 9

Bob, I`d like to thank you alot for your time and help!

I`ll link this forum tread in our forum, so the others can see your work/tips/help.

Much appreciated!

Best regards.

Morten Fredriksen

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**Morten Fredriksen****Member**- Registered: 2016-11-30
- Posts: 9

bobbym wrote:

What's your view on post #1768 and 1769 at page 59 then?

I do not have one. Enderman's work in #1767 which is accurate, is about as far as I can understand.

But I do have an interest in your work. I hike alot and sometimes out of necessity those hikes are in the dark. To a predator, night is the equivalent to dusk for a human. Meaning he can see pretty well. I am totally blind in that environment so a good flashlight is mandatory.

You are very welcome to join our community!

Morten

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**Morten Fredriksen****Member**- Registered: 2016-11-30
- Posts: 9

Bob, what if the LED that will be used in this flashlight has an beam angle of 115 degrees?

In my "eyes" that would not be an efficient LED to us because of that beam angle, it sure would be a deep reflector to catch the light from it? If we would like to have as less "spill" as impossible....

Am I correct?

MF.

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**Morten Fredriksen****Member**- Registered: 2016-11-30
- Posts: 9

If you would like to throw in som more variables in this, I belive it would make me feel more dizzy.... lol

Feel free: www.desmos.com/calculator/12vky2fcgl

*Last edited by Morten Fredriksen (2016-12-01 10:24:26)*

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,000

hi Morten,

The angle of the LED isn't the critical factor. In this diagram I've shown two rays escaping from the sides of the mirror. Ideally, you want all light from the LED to hit the reflector so you get maximum light sent forward.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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