Coordinates for the lower curve: Coordinates for the upper curve:
1.2. Sine regression
After listing these coordinates into the list function on my GDC, a Texas TI-84 Plus, I can use the sine regression to find an appropriate function for the points.
The lower curve (g(x)):
y = a ∙ sin ( bx + c ) + d
a = 1.668240274
b = 0.8100728241
c = –2.216031762
d = 1.96525663
g(x) = 1.668240274 sin ( 0.8100728241 x – 2.216031762 ) + 1.96525663
The upper curve (f(x)):
y = a ∙ sin ( bx + c ) + d
a = 2.378686945
b = 0.7859285241
c = –2.565579771
d = 3.202574671
f(x) = 2.378686945 sin ( 0.7859285241 x – 2.565579771 ) + 3.202574671
By comparing the two functions to the points of the logo, it can be noticed that these functions do not represent the two curves in the logo very precisely. This can be explained by that the curves are not sine functions. Evidence for this is that for a function to be a sine function it has to be symmetric, and the curves of the logo do not have that property.
1.3. Cubic regression
Another type of function that may represent the curves can be polynomial function of 3rd order, a cubic function. Using the graphing package, “Graph 4.3” by Ivan Johansen downloaded at , I can find the polynomial function that best represent the curves.
The lower curve (g(x)):
y = ax3 + bx2 + cx + d
a = –0.089515466
b = 0.74847837
c = –0.88572438
d = 0.58866081
R2 = 0.9995
g(x) = –0.089515466x3 + 0.74847837x2 – 0.88572438x + 0.58866081
The upper curve (f(x)):
y = ax3 + bx2 + cx + d
a = –0.13020708
b = 1.2745508
c = –2.4169654
d = 2.0480192
R2 = 0.9995
f(x) = –0.13020708x3 + 1.2745508x2 – 2.4169654x + 2.0480192
By trial and error, a more simple form of the functions can be discovered. By changing the level of accuracy to 4 decimal places, a simpler form of the functions can be found.
g(x) = –0.0895x3 + 0.7485x2 – 0.8857x + 0.5887
f(x) = –0.1302x3 + 1.2746x2 – 2.4170x + 2.0480
1.4. Comparing the sine and the cubic regression
To show graphically that the cubic functions represent the logo better than the sine functions, we can place them both in the same coordinate diagram and compare them.
We can see that the cubic functions represent the points of the logo much better than the sine functions, so the best way to represent the logo with mathematical functions is to use cubic functions.
There are limitations to these calculations; the main limitation may be inaccuracy of measurements of the points.
In addition, a regression line appears to fit the central portion of the scatter plot well, so there will always be an uncertainty to how well the line represents the plots.
The R2 value is statistic variables that represent the coefficient of determination, and it represents the accuracy of the graph according to the points. Both of the R2 values given from my calculations were the same, 0.9995, this means that the limitations for both graphs are the same, and by that it can be concluded that the mathematical model represent the logo very well.
2. Modifying the logo to double dimensions for t-shirts
2.1. Modifying the logo to double dimensions
The functions have to be modified for the curves to fit into double dimensions. In double dimensions, the axes are twice the length as in normal dimensions.
x-axis = 13.0
y-axis = 12.2
For the graphs to fit in these new dimension they have to be stretched both vertically and horizontally.
For a function to be vertically stretched, it has to be multiplied by a factor of r. The expression for a vertical stretch of the function will be:
rf(x) = r(ax3 + bx2 + cx + d)
For a function to be horizontally stretched, the x-value has to be multiplied by
The expression for a horizontal stretch of the function will be:
fx = ax3 + bx2 + cx + d
As the dimensions are doubled, the value of r will be 2, and the expression of the modified function will be:
2 f(½x) = 2( a(½)x3 + b(½)x2 + c(½)x + d)
g(x) = 2(–0.0895(½)x3 + 0.7484(½)x2 – 0.8857(½)x + 0.5886)
f(x) = 2(–0.1302(½)x3 + 1.2745(½)x2 – 2.4169(½)x + 2.0480)
3. Modifying the logo for a buisines card
3.1. Modifying the logo
If this logo was to be printed to a standard business card of 9 by 5 cm the functions also have to be modified.
To calculate the value for the vertical and horizontal stretch, the new dimensions have to be divided by the original dimensions. The values calculated will represent the stretches.
Vertical stretch: 5/6.1 = 0.82
Horizontal stretch: 9/6.5 = 1.38 1/1.38= 0.72
g(x) = 0.82 (-0.0895(0.72)x3 + 0.7484(0.72)x2 – 0.8857 (0.72)x + 0.5886)
f(x) = 0.82 (-0.1302(0.72)x3 + 1.2745(0.72)x2 – 2.4169 (0.72)x + 2.0480)
3.2. The area fraction of the logo
To find out what fraction of the area of the card the logo occupy, the integral of the lower curve has to be subtracted from the integral of the upper curve. The result of the integral calculations then has to be divided by the total area of the card.
=
=
An accurate way to calculate this is to use the integral function on the GDC.
This gives the area(A) beneath the two curves to be:
A(g(x)) = 15.5419
A(f(x)) = 23.6114
area = 23.6114 – 15.5419
= 8.0695
Total area of the card: 9∙5 = 45
= 0.179322
It is impossible to simplify this fraction, but if we first simplify the answer to fewer decimals it is possibe to get a simplified fraction.
0.18 =
So the logo occupy of the card surface.
This might be an important aspect of a business card because it is important for a logo to be noticed but at the same time not take up to much space.
4. Conclusion
In the introduction I asked if the area of the logo on the business card was the same as the area of the original logo, and to answer this the area of the initial logo has to be calculated.
Using the GDC the area of the initial logo is given to be:
0.18 =
So the answer is yes, the area of the logo on the business card is the same as the area of the original logo. This means that all the calculations and modifications are correct and that the logo is precicely modified into other dimensions.
