Finding the Pitch of a Log Spiral: Solving the Equation and Graphing the Results

[bobie]

Homework Statement

This is not homework,
I'd like to know if it is possible to find the pitch a log spiral knowing that at every round (2π) r grows by 535.5 times.

For the little I know the equation should be just r = e^2π = 535.5
Supposing it starts at r=1cm and goes clockwise, I have guessed that its length should be 755.88 cm. and pitch=45° ,
but according to Torricella formula pitch should be ≈44.85°, since the whole length should be ≈ 759 cm and \varphi= ≈45.15°
so that 535.3/cos\varphi =759 cm and 759-755.9 = 3 cm which is too much for the curve from 1 to 0, isn't it?

Can anyone make a graph?

Thanks a lot.

 

[BvU]

Not homework, so no template ? I am but human, so I need some more givens!

A circular helix (fixed radius 3D spiral) has a pitch: the width of one complete helix turn, measured parallel to the axis of the helix.

I take it you mean a different pitch, like here. But if you found that, you have also found the expression for the pitch, right?

And what angle has a tangent 1 ?

A picture of one turn looks like an ear on it side. Any subsequent turn is almost completely similar (preceding turns are so small they disappear in the dot at the origin).

 

[Simon Bridge]

iirc, pitch is:
$\varphi = \frac{\pi}{2}-\arctan(\frac{1}{b})$ where $b$ comes from: $r=ae^{b\theta}$

if every $2\pi$ radiens, $r$ grows by $s$ times, then $r(\theta+2\pi)=sr(\theta)$

$$s = \frac{r(\theta+2\pi)}{r(\theta)} = \frac{ae^{b(\theta+2\pi)}}{ae^{b\theta}} = e^{2\pi b}\\

\qquad \implies b=\frac{\ln s}{2\pi}\\

\qquad \implies \varphi = \frac{\pi}{2} - \arctan \left( \frac{2\pi}{\ln s} \right)$$

... but like BvU, cannot comment further without more information.

 

[bobie]

here. But if you found that, you have also found the expression for the pitch, right?
And what angle has a tangent 1 ?
.

Thanks, BvU & Simon, I found that expression

\arccos \frac{\langle \mathbf{r}(\theta), \mathbf{r}'(\theta) \rangle}{\|\mathbf{r}(\theta)\|\|\mathbf{r}'(\theta)\|} = \arctan \frac{1}{b} = \phi.

at wiki alright, but (not excluding I am so dumb I can't use it), I am a poor student and my pocket calculator has no arctan etc.. function.

iirc, pitch is:
$\varphi = \frac{\pi}{2}-\arctan(\frac{1}{b})$ where $b$ comes from: $r=ae^{b\theta}$

b is 1, arctan 1 = 45°, that need no calculator but I was not sure about b!

I had reached that conclusion by logic, and also made a graph and saw the angle (pitch) is rougly 45°,
but, as I explained, the Torricelli formula for the whole curve led me astray.

the limit as \theta goes toward -\infty is finite. This property was first realized by Evangelista Torricelli even before calculus had been invented.[4] The total distance covered is \textstyle\frac{r}{\cos(\phi)}, where r is the straight-line distance from P to the origin.

So both pitch and \varphi are exactly 45°?

Thanks again!

 

[Simon Bridge]

Scientific calculators will have an "inv" or "2nd" button, pressing that and the "tan" button gets you the arctan.
http://web2.0calc.com/ ... for instance, used the "2nd" button.

If I put s=535.5, then, using GNU/Octave:

> (pi/2 - atan(2*pi/log(535.5)))*180/pi
ans =  45.000

... yep, the answer is, pitch is exactly 45deg or pi/4 radiens.
It is usually a good idea to think in radiens for angles.

Aside: in my equation I assigned the symbol $\varphi$ to pitch - possibly confusingly.
I'd have written $\varphi+\phi = \frac{\pi}{2}$ where $\phi$ is the usual characteristic angle between the tangent and the radial line.

 

[bobie]

... yep, the answer is, pitch is exactly 45deg or pi/4 radiens.
It is usually a good idea to think in radiens for angles.
.

So pitch is π/4, what is the length of the curve from 1 to 535.3 (2π)? what a bout Torricella?

 

[Simon Bridge]

for 1<r<535.3 units, the length around the curve would require computing the line integral.
what do you need these exact figures for?

 

[bobie]

for 1<r<535.3 units, the length around the curve would require computing the line integral.
what do you need these exact figures for?

I d' like to check how good is my logic, as I derived the exact result by astract considerations:
L _1,2π = 755.88535482
If we use Torricella to find the whole length and then subtract the remaining curve from 1 to 0, do we get a good result?
If you can't be bothered to calc the integral, can you tell me just the formula? I'll try at Wolframalpha
Thanks, anyway for your kind attention, Simon.

 

[Simon Bridge]

Well, I may be bothered to do the calculation if I knew what it was for.
Writing out the integral equation would amount to doing the calculation.
Do you not know how to do line integrals?

 

[BvU]

iirc

must be something like the screeching of the brake and the smell of burning rubber. My lawyer says I didn't write anything wrong , but I am the first to admit I overlooked the complement thing and did as if $\phi$ itself were the pitch...

2: bobie is leading us on saying he is a poor student. Poor students don't have computers where you can type atan(1) in the googline and get 0.785398163 . Poor students have 1.59$ scientific calculators that even give one digit more! How about that for "exactness"!

3: corroborated by "abstract considerations" leading to L _1,2π = 755.88535482 being 11 digits of precision. No calculator ? Rotfl !

 

[BvU]

is GNU clearsighted or just single precision fuzzy ? ?
$e^{2\pi} = 535.4916555247...$

Ah, no, I spy with my googline eye that atan(2*pi()/ln(535.5)))*180/pi() gives 44.999929 and GNU is off the hook.

Over to our dennis the menace lookalike:
Houston, we have a problem: googline (exp(2*pi)-1)*sqrt(2) = 755.885348218 , a whopping 7 x 10^-6 smaller than the abstract considerations result. Or a typo. On your part.

Became nervous: perhaps GNU is fine and Zuckerberg zucks (can't resist the pun). But no, Excellent spreadsheet (known to be double precision) backs Suckerberg!

So, naughty Bo: Spit it out: what's the magic abstract consideration other than a plagiarism of Evangelista's brilliant work ?

 

[bobie]

... what's the magic abstract consideration other than a plagiarism of Evangelista's brilliant work ?

Hi BvU, I suppose your posts are funny, but they are incomprehensible to a poor student whose language is not English. I don't know GNU or what you mean by 0.785398... or 44.999...
It's good for you to be suspicious, it helps in life, but 2 weeks ago I already guessed the speed of the and must be √2, since it's going at 1cm/s and it's being pushed by the rubber at 1cm/s in the normal direction.
I opened this thread because a strict HWH said I was wrong, led me astray and refused to further collaborate. I sought comfort here.!

...The problem is slightly different since the bullet has inertia radially but there is no rubber rope to pull it tangentially toward car B; the effective speed is > √2v

My guess is ≈24.3 cm

(time 17.2* v √2 = 24.3)
Time (e^2π-1)* velocity (√2) = (on a 10-digit 10-y-old 10$ calculator) gives that result (with an intrusive 5 that drove you round the bend )
I owed you this, since you have been more cooperative.

If you want to show me how to get that result by a line integral, I'll learn one more thing,
dennis

 

[BvU]

Makes two of us: I do try to keep a sniff of humor in my slow typings, my first language isn't english, and I don't know GNU either. Chet seems to know how to deal with this beast, so please read "Chet's calculator". My calculators are either Excel or the Google search textbox.

So on a Google search screen:

• Search for atan(1) and you get 0.785398163
• Searching for pi()/4 gives 0.78539816339 and funny enough it then also shows a scientific calculator, don't ask me why or whether it is reproducible.
• Search atan(2*pi()/ln(535.5))*180/pi() and get 44.999929 rad [the rad is wrong, but understandable]
• Search (exp(2*pi)-1)*sqrt(2) and get 755.885348218
  Ah, the calculator appears after the * in exp(2*... so by then it knows I need a function result
  No extra 5 there ! Maybe your calculator is losing bits and needs lubrication..
  is an new smiley for me! Know any others that can help me spice up PF?

Your typing is so fast my reading can't always keep up (weeks ago I already guessed the speed of the and must be √2 ?? and bullets and cars appear on stage ?? In a fight with a HWH ?? - was that a HWHWH or a LWHWH - HeavyWeight or LightWeight ??

All HWHs are volunteers I should think, doing the best they can; they don't deserve being grabbed by the throat by the screwballs they try to rescue from drowning in ignorance! I note that even I get carried away a bit. You're really living up to your icon picture example! Back to business:

the 17.2 and the guessed 24.3 come out of the blue too as far as I can read. The other post where the dust on battlefield is still settling?

Line integral is ET's work, he's going to $-\infty$. Is that want you want to see worked out?
I mean for a curve $r = r_0 e^\theta$ the arc length
$$ \int_0^{2\pi} ds \ = \ \int_0^{2\pi} \sqrt{x^2+y^2} \, d\theta \ \ {\rm when}\ \ x = r_0 e^\theta \cos \theta {\rm \ \ and\ \ } y = r_0 e^\theta \sin \theta $$ can't be the problem, can it ?

(Stil trying to untangle the OP, notice you suppose it goes clockwise. Clockwise $r_0 e^\theta$ means inward in my perception... but then its arc length is really small)

 

[Simon Bridge]

The scientific calculator widget is an interface for the program that google used to do the maths, it should pop up with all calculations you actually search for JIC you want to follow up on the first result.

You can get arctangent on that calculator by pressing "inv" then "tan" on the buttons.
Anyone with a browser can do it. You have to enter the argument after the function.

Access to PF would normally indicate access to a browser.