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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 |
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*[[Signals and systems|Signals and Systems]] |
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==An Introduction to the Fourier Transform== |
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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]] |
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<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. |
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<br>For example, a square wave could be represented by: |
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<math>x_{\mathrm{square}}(t) = \frac{4}{\pi} \sum_{k=1}^\infty {\sin{\left ((2k-1)2\pi ft \right )}\over(2k-1)} </math> |
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'''Data:''' |
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That's alot of numbers seemingly out of the blue, at first observance. I bet your questions are: |
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* Outside tire diameter = 36.5" or up to 40.5" due to tire expansion |
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* Inside tire diameter = 16" |
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<br> |
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=== 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? === |
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<br> |
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=== 2.Can you show me some really easy plug-in formulas so I can get my homework done faster so I can go play World of Warcraft? === |
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No. Stop playing World of Warcraft. It is a horrid waste of time and money.<br> |
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=== 3.Why would I bother getting one? I don't know what to do with it. === |
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* Width of tire = 17" |
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* Air pressure in tire = 7 psi |
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* Volume of Tire = 9.82 ft^3 **(you have to add about 1.5ft^3 to account for possible expansion)** |
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* Fastest quarter mile time = 4.428 sec |
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X(f)=\int_{-\infty}^{\infty} x(t) e^{-j2\pi ft}\, dt |
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* Fastest quarter mile speed = 337.58 |
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</math> |
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<br> |
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<br> |
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Suppose that we have some function, say <math> \beta (t) </math>, that is nonperiodic and finite in duration.<br> |
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This means that <math> \beta(t)=0 </math> for some <math> T_\alpha < \left | t \right | </math> |
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<br><br> |
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Now let's make a periodic function |
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<math> |
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\gamma(t) |
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</math> |
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by repeating |
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<math> |
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\beta(t) |
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</math> |
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with a fundamental period |
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<math> |
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T_\zeta |
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</math>. |
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Note that |
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<math> |
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\lim_{T_\zeta \to \infty}\gamma(t)=\beta(t) |
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</math> |
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<br> |
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The Fourier Series representation of <math> \gamma(t) </math> is |
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<br> |
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<math> |
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\gamma(t)=\sum_{k=-\infty}^\infty \alpha_k e^{j2\pi fkt} |
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</math> |
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where |
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<math> |
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f={1\over T_\zeta} |
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</math> |
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<br>and |
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<math> |
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\alpha_k={1\over T_\zeta}\int_{-{T_\zeta\over 2}}^{{T_\zeta\over 2}} \gamma(t) e^{-j2\pi kt}\,dt |
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</math> |
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<br> |
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<math> \alpha_k </math> can now be rewritten as |
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<math> |
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\alpha_k={1\over T_\zeta}\int_{-\infty}^{\infty} \beta(t) e^{-j2\pi kt}\,dt |
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</math> |
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<br>From our initial identity then, we can write <math> \alpha_k </math> as |
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<math> |
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\alpha_k={1\over T_\zeta}\Beta(kf) |
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</math> |
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<br> and |
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<math> |
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\gamma(t) |
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</math> |
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becomes |
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<math> |
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\gamma(t)=\sum_{k=-\infty}^\infty {1\over T_\zeta}\Beta(kf) e^{j2\pi fkt} |
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</math> |
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<br> |
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Now remember that |
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<math> |
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\beta(t)=\lim_{T_\zeta \to \infty}\gamma(t) |
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</math> |
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and |
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<math> |
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{1\over {T_\zeta}} = f. |
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</math> |
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<br> |
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Which means that |
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<math> |
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\beta(t)=\lim_{f \to 0}\gamma(t)=\lim_{f \to 0}\sum_{k=-\infty}^\infty f \Beta(kf) e^{j2\pi fkt} |
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</math> |
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<br> |
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Which is just to say that |
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<math> |
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\beta(t)=\int_{-\infty}^\infty \Beta(f) e^{j2\pi fkt}\,df |
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</math> |
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<br> |
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<br> |
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So we have that |
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<math> |
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\mathcal{F}[\beta(t)]=\Beta(f)=\int_{-\infty}^{\infty} \beta(t) e^{-j2\pi ft}\, dt |
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</math> |
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<br> |
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Further |
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<math> |
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\mathcal{F}^{-1}[\Beta(f)]=\beta(t)=\int_{-\infty}^\infty \Beta(f) e^{j2\pi fkt}\,df |
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</math> |
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'''Equations:''' |
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==Some Useful Fourier Transform Pairs== |
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Volume = [(pi)*(R^2)(h)-(pi)*(r)(h)] |
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<math> |
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\mathcal{F}[\alpha(t)]=\frac{1}{\mid \alpha \mid}f(\frac{\omega}{\alpha}) |
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</math> |
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<br> |
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{| |
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|- |
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|<math>\mathcal{F}[c_1\alpha(t)+c_2\beta(t)]</math> |
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|<math>=\int_{-\infty}^{\infty} (c_1\alpha(t)+c_2\beta(t)) e^{-j2\pi ft}\, dt</math> |
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|- |
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| |
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|<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> |
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|- |
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| |
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|<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> |
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|- |
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|} |
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<br> |
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<math> |
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\mathcal{F}[\alpha(t-\gamma)]=e^{-j2\pi f\gamma}\Alpha(f) |
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</math> |
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<br> |
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<math> |
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\mathcal{F}[\alpha(t)*\beta(t)]=\Alpha(f)\Beta(f) |
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</math> |
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<br> |
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<math> |
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\mathcal{F}[\alpha(t)\beta(t)]=\Alpha(f)*\Beta(f) |
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</math> |
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<br> |
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Some other usefull pairs can be found here: [[Fourier Transforms]] |
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y = .075/(Vmax-Vmin) **(this is the number of inches that the tire should expand given the current velocity.)** |
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==A Second Approach to Fourier Transforms== |
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*[[Fourier Transforms]] |
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Latest revision as of 21:23, 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.)**