An alternative method of calculating loudspeaker box volume. The only information needed is the nominal driver size.
The program listing shown below can be copy and pasted into the editor window of the BASIC interpreter. Click the "Run" button to execute the program. Width, height, depth values shown are the interior dimensions. No adjustment for the displacement of the driver is needed. Dv is the nominal vent diameter. Multiply Dv by 0.75 for dual vent tubes.
The back and sides of the enclosure interior should be covered with acoustical cotton batts or some similar damping material. Consider rounding the backside of the cutout hole to allow air to escape from the backside of the driver.
20 rem Loudspeaker Box Volume Calculator 30 rem Copyright (c) Donald A. Swanson 40 clear 50 home 60 data 0.904,1.015,1.136 70 input "Enter nominal driver size (in) = ";dz 80 for i = 1 to 3 90 rem calculate width (wd), height (ht), depth (dp) 100 read s1 110 ds = dz * s1 120 wd = ds * 1.618 130 ht = wd * 1.618 140 dp = wd * (1 - (1 / (1 + 5))) 150 wd = int((wd * 4) + 0.5) / 4 160 ht = int((ht * 4) + 0.5) / 4 170 dp = int((dp * 4) + 0.5) / 4 180 rem calculate enclosure volume (Vb) 190 vb = wd * ht * dp 200 rem convert cubic inches to liters 210 vb = vb / 61.024 220 vb = int((vb * 10) + 0.5) / 10 230 rem calculate nominal vent diameter (Dv) 240 dv = ds / 3 250 dv = int((dv * 8) + 0.167) / 8 260 print"---------------------------------------" 270 print 280 print " ";vb;" (L)";" Dv: ";dv;" (in)" 290 print 300 print " ";"w,h,d: ";wd;", ";ht;", ";dp;" (in)" 310 print 320 next i 330 restore 340 print 350 input " Hit enter to continue ";en$ 360 goto 40
A subjective method of making odds from the analysis of thoroughbred horse races.
The leftmost column in figure 1 shows the seven symbols used in the method. The left side symbol part (+, N, Ø) is the base. The right side symbol part (+, -) is the modifier. Symbols are usually hand-written with a forward slash "/" as a separator, for example: N+ / + / Ø+. Symbol selection starts with consideration of the base symbols before moving up or down. Symbols and descriptions are to be memorized. Column 3 in figure 1 shows the matching Fibonacci numbers which will be used to calculate the weighted percentages.
| Symbol | Description | Weight | Confidence Level |
|---|---|---|---|
| ++ | double plus | 21 | maximum |
| + | plus | 13 | high |
| N+ | neutral plus | 8 | moderate high |
| N | neutral | 5 | moderate |
| N- | neutral minus | 3 | moderate low |
| Ø+ | doubtful plus | 2 | low |
| Ø | doubtful | 1 | minimum |
The race is analyzed to decide which horses will be contenders. The handicapper will have some level of confidence that one of the contenders will win the race. A symbol is selected to represent the confidence level. The rightmost column in figure 1 shows the matching confidence levels which are to be memorized. Horses that don't make the cut are labeled as doubtful or Ø.
cp - contender percentage
nc - number of contenders
fs - field size
wt - weight
cp = (wt * nc) / ((wt * nc) + (fs - nc))
Four contenders in a nine horse field with moderate confidence.
4 / 9 / N
N / N / N / N / Ø / Ø / Ø / Ø / Ø
5 + 5 + 5 + 5 + 1 + 1 + 1 + 1 + 1
cp = 20 / 25 = 0.80
cp = (5 * 4) / ((5 * 4) + (9 - 4)) = 0.80
One contender in a six horse field with maximum confidence.
1 / 6 / ++
++ / Ø / Ø / Ø / Ø / Ø
21 + 1 + 1 + 1 + 1 + 1
cp = 21 / 26 = .808
cp = (21 * 1) / ((21 * 1) + (6 - 1)) = .808
Races with two or more contenders will use one to three factors which are to be arranged from left to right in order of importance, for example: class, distance, form. Factors can be a two-part compound, for example: form-surface, jockey-trainer, form-age, distance-surface.
The N symbol is extended to NN making the neutral base symbols N+, NN, N-. The factor symbols are the right side symbol parts (+, N, -) matched up with Fibonacci numbers 8, 5, 3 respectively. Figure 2 shows the six factor symbol combinations and weighted percentages. A symbol combination is chosen that indicates the relative importance of each factor. The entire contents of figure 2 are to be memorized.
Weighted percentage calculation for the "plus and minus" (+ N -) combination.
8 + 5 + 3 = 16
8 / 16 = .5
5 / 16 = .3125
3 / 16 = .1875
Factors can be doubled, for example: speed, speed, distance making a two factor race. The factor symbol representing the doubled factor must be duplicated in the combination as shown in figure 2. A single factor can be tripled (N N N) making a one factor race.
| Symbols | Description | Percent | Doubled |
|---|---|---|---|
| N N N | neutral and neutral | 33,33,34 | 66,34 |
| + N N | plus and neutral | 44,28,28 | - |
| N N - | neutral and minus | 38,38,24 | 76,24 |
| + N - | plus and minus | 50,31,19 | - |
| + - - | two minus | 56,22,22 | 56,44 |
| + + - | two plus | 42,42,16 | 84,16 |
Two contenders in an eight horse field with high confidence.
2 / 8 / +
cp = (13 * 2) / ((13 * 2) + (8 - 2)) = .813
Three factors are chosen: class, distance, form with the "plus and minus" (+ N -) combination.
Calculate the factor percentages.
8 + 5 + 3 = 16
(8 / 16) * cp = .407
(5 / 16) * cp = .254
(3 / 16) * cp = .152
Each contender will get three symbols with one for each factor as shown in figure 3 second column.
| Horse | cls / dst / form | Weights |
|---|---|---|
| #1 | + / N / N | 13, 5, 5 |
| #2 | N- / N+ / + | 3, 8, 13 |
Calculate the weighted percentages for each contender and factor.
13 + 3 = 16
(13 / 16) * .407 = .331 (#1)
(3 / 16) * .407 = .076 (#2)
5 + 8 = 13
(5 / 13) * .254 = .098 (#1)
(8 / 13) * .254 = .156 (#2)
5 + 13 = 18
(5 / 18) * .152 = .042 (#1)
(13 / 18) * .152 = .110 (#2)
Sum the three percentages for each contender.
.331 + .098 + .042 = .471 (#1)
.076 + .156 + .110 = .342 (#2)
Sort the percentages (pct) in descending order before converting into odds.
odds = (1 / pct) - 1
(1 / .471) - 1 = 1.123 (#1)
(1 / .342) - 1 = 1.924 (#2)
Select the increment (inc) for rounding off.
if odds < 0.3 then inc = 20 elseif odds < 1 then inc = 10 elseif odds < 2 then inc = 5 elseif odds < 5 then inc = 2 else inc = 1
odds = int ((odds * inc) + .5) / inc
int ((1.123 * 5) + .5) / 5 = 1.2 (#1)
int ((1.924 * 5) + .5) / 5 = 2 (#2)
Convert rounded odds into fractions.
n = 0.5 // numerator d = 0 // denominator while int(n) <> n d = d + 1 n = odds * d end print n,"/",d
Finished calculation shown in figure 4 with percentages and fractional odds.
| 2 / 8 / + | cls / dst / form | + N - | |
|---|---|---|---|
| #1 | + / N / N | .471 | 6 / 5 |
| #2 | N- / N+ / + | .342 | 2 / 1 |
Figure 5 is a two factor race with the distance-surface factor doubled.
Note that the symbols for each contender are duplicated with the doubled factor.
| 2 / 10 / + | cls / dst-sur | + - - | |
|---|---|---|---|
| #1 | N+ / N / N | .552 | 4 / 5 |
| #2 | N- / Ø+ / Ø+ | .213 | 7 / 2 |