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For this homework assignment I wanted to try and see if I could find a correlation between a dragsters rear tire expansion in comparison to its velocity by using either the method of Laplace transform or the Fourier series. To help me with my model I will me using the Army's dragster for some of my data. If you would like to check it out you can find it at [http://www.goarmy.com/army-racing/nhra-top-fuel/dragster.html]. If you want to watch a video of dragster tires click here : http://www.youtube.com/v/V3yj_OGezWc?version=3
*[[Signals and systems|Signals and Systems]]
==An Introduction to the Fourier Transform==


Unfortunately, the Fourier Transform isn't a Transformer. If it was, you would have seen it in the movie that came out lately. [[Image:transformer_roolbar.jpg]]
<br>One way to explain a Fourier Transform is to say it's a bunch of sinusoids added to create a just about any function you want. Another way to describe it is to say it's a way of representing a function in the frequency domain instead of the time domain.
<br>For example, a square wave could be represented by:
<math>x_{\mathrm{square}}(t) = \frac{4}{\pi} \sum_{k=1}^\infty {\sin{\left ((2k-1)2\pi ft \right )}\over(2k-1)} </math>


'''Data:'''
That's alot of numbers seemingly out of the blue, at first observance. I bet your questions are:
* Outside tire diameter = 36.5" or up to 40.5" due to tire expansion


* Inside tire diameter = 16"
1. Telling me the difference between a transformer and a Fourier Transform hasn't helped me finish my assignment. What else can you tell me? <br>
2.How do I get one? <br>
3.If I've got one, what can I do with it?


* Width of tire = 17"


* Air pressure in tire = 7 psi


* Volume of Tire = 9.82 ft^3 **(you have to add about 1.5ft^3 to account for possible expansion)**


* Fastest quarter mile time = 4.428 sec
X(f)=\int_{-\infty}^{\infty} x(t) e^{-j2\pi ft}\, dt


* Fastest quarter mile speed = 337.58
</math>
<br>
<br>
Suppose that we have some function, say <math> \beta (t) </math>, that is nonperiodic and finite in duration.<br>
This means that <math> \beta(t)=0 </math> for some <math> T_\alpha < \left | t \right | </math>
<br><br>
Now let's make a periodic function
<math>
\gamma(t)
</math>
by repeating
<math>
\beta(t)
</math>
with a fundamental period
<math>
T_\zeta
</math>.
Note that
<math>
\lim_{T_\zeta \to \infty}\gamma(t)=\beta(t)
</math>
<br>
The Fourier Series representation of <math> \gamma(t) </math> is
<br>
<math>
\gamma(t)=\sum_{k=-\infty}^\infty \alpha_k e^{j2\pi fkt}
</math>
where
<math>
f={1\over T_\zeta}
</math>
<br>and
<math>
\alpha_k={1\over T_\zeta}\int_{-{T_\zeta\over 2}}^{{T_\zeta\over 2}} \gamma(t) e^{-j2\pi kt}\,dt
</math>
<br>
<math> \alpha_k </math> can now be rewritten as
<math>
\alpha_k={1\over T_\zeta}\int_{-\infty}^{\infty} \beta(t) e^{-j2\pi kt}\,dt
</math>
<br>From our initial identity then, we can write <math> \alpha_k </math> as
<math>
\alpha_k={1\over T_\zeta}\Beta(kf)
</math>
<br> and
<math>
\gamma(t)
</math>
becomes
<math>
\gamma(t)=\sum_{k=-\infty}^\infty {1\over T_\zeta}\Beta(kf) e^{j2\pi fkt}
</math>
<br>
Now remember that
<math>
\beta(t)=\lim_{T_\zeta \to \infty}\gamma(t)
</math>
and
<math>
{1\over {T_\zeta}} = f.
</math>
<br>
Which means that
<math>
\beta(t)=\lim_{f \to 0}\gamma(t)=\lim_{f \to 0}\sum_{k=-\infty}^\infty f \Beta(kf) e^{j2\pi fkt}
</math>
<br>
Which is just to say that
<math>
\beta(t)=\int_{-\infty}^\infty \Beta(f) e^{j2\pi fkt}\,df
</math>
<br>
<br>
So we have that
<math>
\mathcal{F}[\beta(t)]=\Beta(f)=\int_{-\infty}^{\infty} \beta(t) e^{-j2\pi ft}\, dt
</math>
<br>
Further
<math>
\mathcal{F}^{-1}[\Beta(f)]=\beta(t)=\int_{-\infty}^\infty \Beta(f) e^{j2\pi fkt}\,df
</math>


'''Equations:'''
==Some Useful Fourier Transform Pairs==
Volume = [(pi)*(R^2)(h)-(pi)*(r)(h)]
<math>
\mathcal{F}[\alpha(t)]=\frac{1}{\mid \alpha \mid}f(\frac{\omega}{\alpha})
</math>
<br>
{|
|-
|<math>\mathcal{F}[c_1\alpha(t)+c_2\beta(t)]</math>
|<math>=\int_{-\infty}^{\infty} (c_1\alpha(t)+c_2\beta(t)) e^{-j2\pi ft}\, dt</math>
|-
|
|<math>=\int_{-\infty}^{\infty}c_1\alpha(t)e^{-j2\pi ft}\, dt+\int_{-\infty}^{\infty}c_2\beta(t)e^{-j2\pi ft}\, dt</math>
|-
|
|<math>=c_1\int_{-\infty}^{\infty}\alpha(t)e^{-j2\pi ft}\, dt+c_2\int_{-\infty}^{\infty}\beta(t)e^{-j2\pi ft}\, dt=c_1\Alpha(f)+c_2\Beta(f)</math>
|-
|}
<br>
<math>
\mathcal{F}[\alpha(t-\gamma)]=e^{-j2\pi f\gamma}\Alpha(f)
</math>
<br>
<math>
\mathcal{F}[\alpha(t)*\beta(t)]=\Alpha(f)\Beta(f)
</math>
<br>
<math>
\mathcal{F}[\alpha(t)\beta(t)]=\Alpha(f)*\Beta(f)
</math>
<br>
Some other usefull pairs can be found here: [[Fourier Transforms]]


y = .075/(Vmax-Vmin) **(this is the number of inches that the tire should expand given the current velocity.)**
==A Second Approach to Fourier Transforms==
*[[Fourier Transforms]]

Latest revision as of 20:23, 31 October 2010

For this homework assignment I wanted to try and see if I could find a correlation between a dragsters rear tire expansion in comparison to its velocity by using either the method of Laplace transform or the Fourier series. To help me with my model I will me using the Army's dragster for some of my data. If you would like to check it out you can find it at [1]. If you want to watch a video of dragster tires click here : http://www.youtube.com/v/V3yj_OGezWc?version=3


Data:

  • Outside tire diameter = 36.5" or up to 40.5" due to tire expansion
  • Inside tire diameter = 16"
  • Width of tire = 17"
  • Air pressure in tire = 7 psi
  • Volume of Tire = 9.82 ft^3 **(you have to add about 1.5ft^3 to account for possible expansion)**
  • Fastest quarter mile time = 4.428 sec
  • Fastest quarter mile speed = 337.58

Equations: Volume = [(pi)*(R^2)(h)-(pi)*(r)(h)]

y = .075/(Vmax-Vmin) **(this is the number of inches that the tire should expand given the current velocity.)**

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