बासर/ वा रचनेचे गDणती अिध%ान
ूकरण चौथे
बासर/ वा रचनेचे गDणती अअअिधअिधिधिध%ान%ान%ान%ान
MATHEMATICAL CALCULATIONS FOR TRADITIONAL HINDUSTANI BANSURI OF SIX TONE HOLES
Mathematical Calculations for Traditional Hindustani 6 Tone Hole Flute*
Determining approximate finger hole locations for a traditional 6 tone hole flute
(i.e. 1 blow hole and 6 tone holes total 7 holes) is somewhat complicated.
The wavelength of sound produced is determined by the flute tube and tone holes. For a pipe of length L, open at both ends, and ignoring end effects, the wavelength of sound is twice the length of the tube. The frequency produced is given by dividing the speed of sound (345 m/s) at 25 C by the wavelength.
The method described here is to estimate an effective length for a real (cylindrical) tube taking into account end effects, the size of tone holes, etc.
Let’s discuss the basic anatomy of a primitive flute, the geometries that are involved, and name and number the holes.
१८५
The basic flute is constructed of a single tube with a stopper at one end (typically cork.) This discussion will refer to this end as the “Close” end. The close end is also the end which will have the blow hole. The blow hole is the hole the instrumentalist will blow across to generate the sound (it’s the hole the player blows across.) The blow hole is located just beside the stopper.
The other end of the flute is open and will be called the “Open” end.
This discussion will consider a traditional Hindustani flute of 6 tone holes.
The tone holes are the holes in the flute which the player will open or close using his fingers. The order that the holes are numbered for this discussion will be from 1 to 6, where 1 is the hole closest to the open end and 6 is the hole closest to the
* In illustrating examples, the traditional Hindustani flute we are considering is a Safed 4 (F) Basuri of 6 tone holes. * Throughout this discussion, we will be considering the multiplication symbol “x” as “*”.
close end. Other conventions or definitions may apply when learning how to play such an instrument but we’re talking about this in a way to keep it as simple as possible for the following mathematical calculations. Here’s a summary table .
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TRADITIONAL HINDUSTANI 6 TONE HOLE FLUTE FLUTE TERMINOLOGIES
• Tube – The body of the flute • Stopper – The cork stopper which is closing one end of the tube. Tuning the instrument can be done by moving the stopper. • Close End – The end of the tube which has been closed by the stopper in it. • Open End – The end of the tube which is open. • Blow Hole – The hole the player will blow across to make the flute play. • Tone Hole – Holes that the user will “open” or “close” with his fingers when he plays the flute.
There are many dimensions involved with the flute. Here’s a list of geometric parameters that will be used.
१८७
TRADITIONAL HINDUSTANI 6 TONE HOLE FLUTE VARIABLE PARAMETERS OF TRADITIONAL FLUTE
• t - wall thickness of the tube. • Lo - Length of the tube from the center of the blow hole to the open end. • Les - Effective length of the tube at the close end. • Dt - Inside diameter of the tube • Rt - Inside radius of the tube • Lbs - Distance from center of blow hole to stopper • L0 - Actual real distance from center of blow hole to open end. • Le0 - Effective distance from center of blow hole to open end. • Lbt1 - Distance between the blow hole and tone hole 1. • Lbt2 - Distance between the blow hole and tone hole 2. • Lbt3 - Distance between the blow hole and tone hole 3. • Lbt4 - Distance between the blow hole and tone hole 4. • Lbt5 - Distance between the blow hole and tone hole 5. • Lbt6 - Distance between the blow hole and tone hole 6. • Dbh - Diameter of blow hole. • Dt1 - Diameter of tone hole 1.
१८८
• Dt2 - Diameter of tone hole 2. • Dt3 - Diameter of tone hole 3. • Dt4 - Diameter of tone hole 4. • Dt5 - Diameter of tone hole 5. • Dt6 - Diameter of tone hole 6. • Rbh - Radius of the blowhole.
If the fundamental (all tone holes closed) corresponds to an effective length, then the desired effective lengths for a six (or fewer) hole flute to produce notes of the Hindustani (Indian) flute are given by the following table:
Note Note Note actually Ratio, i.e. Number being played length units of Le0 0 Sa Pa 1 1 Re Dha 8/9 2 Ga Ni 4/5 3 Ma(Sudha) Sa 3/ 4 4 Pa Re 2/3 5 Dha Ga 16/27 6 Ni(Teevra) Ma(Teevra) 128/243
The actual physical length of the flute will be shorter than Le0 and the distance to the tone holes (from the close end) will be shorter than the values calculated. The end effects (at both the blow hole and at the first open hole) act (approximately) as an additional length which must be subtracted from the calculated effective lengths to get the physical lengths desired. As long as you don’t deviate too much from typical flute proportions, these approximate calculations should get you pretty close.
Step One
१८९
First, make your flute with no tone holes and adjust the length to match the desired lowest note. You can do this by estimating what the tube length should be by using the following formula.
Tube Length = the wavelength / 2
= (Speed of Sound / Frequency) / 2
For a flute which you would like to be of note ‘Sa’ of safed 4 (F) (Frequency 349.22 Hz) i.e., the open end note is ‘Kharja Pa’ (frequency 261.92 Hz) the calculations are as follows.
If sound travels at 345 * 0.904 meters per second we can estimate the length you should start your tube at.
For our example,
345*0.904 = 311.88
Note: 0.904 is the multiplying factor for L0 open end length corrections at 25 C.
Frequency of Pa (safed 4) = 523.84 Hz
Frequency of Kharja Pa (open end note) = 523.84 / 2 = 261.92 Hz
Tube Length = 311.88 / (2 * 261.92) = 593 mm
You want to add about 80 mm* for the stopper and adjustment of the blow hole. Therefore you should cut your initial tube to 593 + 80 = 673 mm, rounded to 675 mm. Now you want to drill the blowhole.
With reference to page no of part “Basic Principles of Flute” point number 15, the appropriate blow hole diameter for effortless playing/blowing for a Hindustani flute is 11 to 11.5
१९० mm. This enhances the flute’s properties, such as, sound quality, tunefulness and octave range.
If the blow hole is of a bigger diameter, the tone is deeper. However, the power and stamina which needs to be exerted for playing also increases. Therefore, flutes with bigger diameters are not recommended for elaborate musical
*The additional 80 mm length is to keep safe distance of the stopper cork from the end of the tube to avoid easy tampering after setting the same for final tuning. performances lasting for more than 3 hours. Also, octave range reduces to 2 or even more so.
Oval shaped blow holes are good for flute playing.
For our example,
We are going to choose a Bamboo or PVC tube of Inner diameter 25 mm.
As such, we have taken the values of
Example Variable Results Dt 25 mm Dbh 11.5 mm
Step Two
Cut some material (cork or rubber) with a diameter which will snugly fit into the tube. We now have to calculate where the “stopper” will need to be placed. In reality it doesn’t matter too much as long as the inner edge is pretty close to the blowhole.
१९१
We’ll also need to know the thickness of the tube wall. You will have to find this using calipers or use the specifications of the tube.
Example Variable Results t 2 mm
We need to calculate temporary value for further calculations called Heff.
Heff = t + 0.85 * Dbh
For our example,
Heff = 2 + (0.85 * 11.5) = 11.775 mm
Now we’ll define a parameter called e where e = fraction of blow hole covered by the player’s lip and has a value typically in the range of 0 to 0.15 and in some special cases even negative. In this example, we’ll use 0.1 for a value of e. With this value now defined, it’s possible to calculate Les.
Les = Heff * (1-e) * (Dt / Dbh)^2
For our example,
Les = 11.775 * (1-0.1) * (25/11.5) * (25/11.5) = 11.775 * 0.9 * 2.174 * 2.174
= 50.0827 mm
Using a single Les for all notes is a reasonable approximation IF Lbs is close to 0.37 Les. We want to know Lbs so that we know
१९२ where to put the stopper that we made therefore we’ll use the following expression to calculate Lbs.
Lbs = 0.37 * Les
For our example,
Lbs = 0.37 * 50.0827 = 18.531 mm
Step Three
Now you have to adjust the length of the flute by blowing across the blow hole and shortening the length of the tube (cutting or sanding off lengths of the open end) until the tone you get out of the flute corresponds with the note (in this case a Kharja Pancham of F. F Shadja = 349.22 Hz. F Pancham = 1.5 * Sa Frequency = 1.5 * 349.22 = 523.83 Hz. Kharja Pancham = Pancham / 2 = 523.83 / 2 = 261.9 Hz.) that you’re targeting. You will need a frequency meter or tone generator (like a harmonium or other instrument like keyboard) to compare to. The final length that results is very important. After that adjustment has been made, you’re ready to start calculating!
After the flute has been cut to the correct tonal length, get out your ruler and measure this distance from the center of the blowhole to the end of the tube. We call this L0.
It is very essential to be able to judge minute frequencies differences upto 1 shruti for correct length adjustments. This is possible if you use keyboard for checking the high / low shades of the note Kharja Pa.
Example Variable Results L0 575 mm
१९३
We’re now ready to calculate the effective length of the tube from the center of the blowhole to the end of the tube. This is a fictitious length and is used to compensate for end effects which exist in the real instrument. We’re going to call this effective length Le0. The first thing we need is the tube inner radius.
Rt = Dt / 2
For our example, Rt = 25 / 2 = 12.5 mm
Next, we can calculate Le0.
Le0 = Les + L0 + (0.6 * Rt)
For our example,
Le0 = 50.0827 + 575 + (0.6 * 12.5) = 632.5823 mm
Step Four
Now use your value of Le0 and the table describing the individual notes of the scale to get desired effective lengths for each of the hole positions, measured from the center of the blowhole to the center of the tone hole. Call these Lebti where ‘i’ corresponds to the tone hole number (e.g. Lebti = 8 / 9 Le0 etc.)
For our example, let’s calculate the first two.
Lebt1 = 8 / 9 * 632.5823 = 562.296 mm
Lebt2 = 4 / 5 * 632.5823 = 506.0661 mm
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The following table shows the results for all 6 values.
Example Variable Results Lebt1 562.295 mm Lebt2 506.066 mm Lebt3 474.437 mm Lebt4 421.722 mm Lebt5 374.864 mm Lebt6 333.212 mm
Let’s restructure this table a little bit because we’re going to be calculating lots of numbers for each hole. As we keep on adding parameters, the table will grow.
Parameter Hole1 Hole2 Hole3 Hole4 Hole5 Hole6
562.295 506.066 474.437 421.722 374.864 333.212 Lebti mm mm mm mm mm mm Step Five
Next, we’re going to pick our tone hole diameters and calculate thickness corrections for each hole. For our example, we’re going to use the following diameters:
Parameter Hole1 Hole2 Hole3 Hole4 Hole5 Hole6 Dthi 11 mm 11.5 mm 10 mm 11 mm 11.5 mm 10 mm (Note: These diameters are specifically chosen for effortless closure and the possibility of expanding the flute to have more tone holes and thus increasing the octave range of the flute.)
For example
१९५
• Left thumb tone hole (closest hole to blow hole) for fine bridging of 2 consecutive octaves (i.e. glissando or meend). This is the soul of Hindustani classical music. • Right little finger tone hole for adding 2 more notes in the base octave. This increases the range of the flute into the Base octave, but it is also used for playing notes in the twitter and high twitter octave. Moreover, it is used for fine bridging of 2 consecutive octaves.
The actual position of each hole will depend on the tone hole size and the position of any other open tone holes. Hence, you have some leeway to choose one or the other of these. Smaller tone holes will give a mellower sound and larger tone holes give a brighter (and louder) sound. Your holes do not need to be all the same size, so it is possible to make some choices which affect the ergonomics of your flute.
Below we will assume that all the tone holes are fixed in size, and only their positions are to be adjusted. If this results in tone hole positions which are uncomfortable (or unusable) then adjust one or more of the tone hole sizes and recomputed.
Parameter Hole1 Hole2 Hole3 Hole4 Hole5 Hole6 562.295 506.066 474.437 421.722 374.864 333.212 Lebti mm mm mm mm mm mm Dthi 11 mm 11.5 mm 10 mm 11 mm 11.5 mm 10 mm
This provides enough information to calculate the thickness correction lengths.
Ltci = t * (Dthi/Dt)^2 / 4 + (All Ltcl’s)
१९६
Where: j = The number of closed holes between the first open hole and the blow end
The expression is attempting to show the correction factor for a specific hole ‘I’ and includes all the thickness corrections for all holes between hole i and the blow hole. The value of j represents the number of holes which are included in the thickness correction.
Parameter Hole1 Hole2 Hole3 Hole4 Hole5 Hole6 562.295 506.066 474.437 421.722 374.864 333.212 Lebti mm mm mm mm mm mm Dthi 11 mm 11.5 mm 10 mm 11 mm 11.5 mm 10 mm Value of i 1 2 3 4 5 6 Value of j 5 4 3 2 1 0 For our example, we will calculate all of them because this is a confusing parameter. We have to work our way from hole 6 to hole 1. This is because we have to sum all the thickness corrections for each of the tone holes as we work our way from the blow hole to the open end of the flute.
Ltc6 = 0 (There are no holes in between hole number 6 and the blowhole, therefore, no correction is required)
Ltc5 = 2 * (11.5 / 25) * (11.5 / 25) / 4 + (0) = 0.10580 mm
Ltc4 = 2 * (11 / 25) * (11 / 25) / 4 + (0.1058) = 0.2026 mm
Ltc3 = 2 * (10 / 25) * (10 / 25) / 4 + (0.2026) = 0.2826 mm
Ltc2 = 2 * (11.5 / 25) * (11.5 / 25) / 4 + (0.2826) = 0.3884 mm
१९७
Ltc1 = 2 * (11 / 25) * (11 / 25) / 4 + (0.3884) = 0.4852 mm
The following table shows the thickness corrections for each of the holes.
Parameter Hole1 Hole2 Hole3 Hole4 Hole5 Hole6 562.295 506.066 474.437 421.722 374.864 333.212 Lebti mm mm mm mm mm mm 11.5 Dthi 11 mm 11.5 mm 10 mm 11 mm mm 10 mm Value of i 1 2 3 4 5 6 Value of j 5 4 3 2 1 0 0.4852 0.3884 0.2826 0.2026 0.1058 Ltci mm mm mm mm mm 0 Ultimately, we’re also going to need the thickness correction factor when all the tone holes are open. This will be the thickness correction for the entire tube length and corresponds to a value of j = 6.
For our example,
Ltc0 = 2 * (10/25) * (10/25) /4 + 0.4852 = 0.5652 mm
Step Six
We’re now going to compute the final lengths for the flute starting with the final tube length and then working our way from tone hole 1 to tone hole 6. The first step is to compute the final tube length.
L0f = L0 – Ltc0
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For our example,
L0f = 575 – 0.5652 = 574.4348 mm
You should now cut of your tube for a final time. The length L0f is the distance between the center of the blowhole to the open end of tube.
Step Seven
The next is to compute the length for hole number 1. Computations of hole length require iteration. This means that the computation will need to be repeated several times substituting some results from a previous step into the new step until the results stop changing. This may seem complicated at first but the example will show you how to do it.
We need a jump off point so we’re going to calculate the initial length using the following formula.
L0bti = Lebti – Les – (0.6 * Rt)
The expression takes into account the correction for the length at the close end and the value of 0.6 * Rt is the length correction at the close end.
For our example,
L0bt1 = 562.295 – 50.0827 – (0.6 * 12.5) = 504.713 mm
१९९
The first hole uses the following formulas.
K1 = 0.75 * Dth1 + t
K2 = (Dth1 / Dt) ^2
D = L0f – Lbt1
Lc1 = K1 / (K2 + (K1 / D))
For our example, we’ll calculate K1 and K2 which don’t change throughout the iterative process.
K1 = 0.75 * 11 + 2 = 10.25 mm
K2 = (11 / 25) * (11 / 25) = 0.1936
Parameter Hole1 Hole2 Hole3 Hole4 Hole5 Hole6 562.295 506.066 474.437 421.722 374.864 Lebti mm mm mm mm mm 333.212 mm 11.5 11.5 Dthi 11 mm mm 10 mm 11 mm mm 10 mm Value of i 1 2 3 4 5 6 Value of j 5 4 3 2 1 0 0.4852 0.3884 0.2826 0.2026 0.1058 Ltci mm mm mm mm mm 0 10.25 K1 mm K2 0.1936 Now we start the iteration process. The iterative formula is as follows.
Lbti(new) = Lebti – Les – Lci - Ltci
We’ll start the first iteration for hole 1.
२००
D = 574.4348 - 504.713 = 69.7218 mm
Lc1 = 10.25 / (0.1936 + (10.25 / 69.7218)) = 30.093 mm
Lbt1(new) = 562.295 - 50.0827 – 30.093 – 0.4852 = 481.6341 mm
Second Iteration; use the last version of Lbt1(new) to recompute D. Proceed with the iteration.
D = 574.4348 - 481.6341 = 92.8007 mm
Lc1 = 10.25 / (0.1936 + (10.25 / 92.8007)) = 33.711 mm
Lbt1(new) = 562.295 - 50.0827 – 33.711 – 0.4852 = 478.0161 mm
Continue the process until Lbt1(new) stops changing. The final result will be Lbt1 which is the actual length which should be used to drill hole 1.
The final result after 6 iterations is 477.4841 mm.
Step Eight
The calculation for the next 5 holes will use a different set of formulas than for hole number 1 but the steps are very similar. Here are the formulas.
२०१
K1 = 0.75 * Dthi + t
K2 = (Dt / Dthi) ^2
Note: K2 for the tone holes other than 1 has an inverted ratio than hole 1.
S = (Lbt(i-1) – Lbti) / 2
Lc1 = S * [(1 + 2 * K1 * K2 / S) ^1/2 - 1]
Let’s work on hole number 2. We’ll calculate K1 and K2 which don’t change throughout the iterative process.
K1 = 0.75 * 11.5 + 2 = 10.625 mm
K2 = (25 / 11.5) * (25 / 11.5) = 4.726
Because the tone hole diameters are all different for our example problem, K1 and K2 are different for each of the holes 2 to 6.
Parameter Hole1 Hole2 Hole3 Hole4 Hole5 Hole6 562.295 506.066 474.437 421.722 374.864 333.212 Lebti mm mm mm mm mm mm Dthi 11 mm 11.5 mm 10 mm 11 mm 11.5 mm 10 mm Value of i 1 2 3 4 5 6 Value of j 5 4 3 2 1 0 0.4852 0.3884 0.2826 0.2026 0.1058 Ltci mm mm mm mm mm 0 10.25 10.625 10.25 10.625 K1 mm mm 9.5 mm mm mm 9.5 mm K2 0.1936 4.726 6.25 5.165 4.726 6.25 Lbti 477.4841
२०२
We need the jump off point to start the iteration process just like we did for step seven.
For our example,
L0bt2 = 506.066 - 50.0827 – (0.6 * 12.5) = 448.4835 mm
Now, we’ll start the iterative process using the same iterative formula as in step seven.
We’ll start the first iteration for hole 2.
S = (477.4841 – 448.4835) / 2 = 14.5003 mm
Lc2 = 14.5003 * [(1 + 2 * 10.25 * 4.726 / 14.5003) ^1/2 - 1]
= 25.6879 mm
Lbt2(new) = 506.066 - 50.0827 - 25.6879 – 0.3884 = 429.907 mm
Second iteration; use the last version of to recompute S. Proceed with the iteration.
S = (477.4841 – 429.907) / 2 = 23.78855 mm
Lc2 = 23.78855 * [(1 + 2 * 10.25 * 4.726 / 23.78855) ^1/2 - 1]
२०३
= 29.7894 mm
Lbt2(new) = 506.066 - 50.0827 - 29.7894 – 0.3884 = 425.805 mm
Continue the process until Lbt2(new) stops changing. The final result will be Lbt2 which is the actual length which should be used to drill hole 2.The final result after 6 iterations is 424.026 mm.Repeating the same iterative procedure for holes 3 thorough 6 will provide the final hole positions.
२०४
Parameter Hole1 Hole2 Hole3 Hole4 Hole5 Hole6
562.295 506.066 474.437 421.722 374.864 333.212 Lebti mm mm mm mm mm mm
Dthi 11 mm 11.5 mm 10 mm 11 mm 11.5 mm 10 mm
Value of i 1 2 3 4 5 6
Value of j 5 4 3 2 1 0
0.4852 0.3884 0.2826 0.2026 0.1058 Ltci mm mm mm mm mm 0
10.25 10.25 10.25 10.25 10.25 K1 mm mm mm mm 10.25 mm mm
K2 0.1936 4.726 6.25 5.165 4.726 6.25
477.484 424.026 394.479 338.157 294.932 249.064 Lbti mm mm mm mm mm mm
Ratio of lengths * 0.83 0.737 0.686 0.588 0.513 0.433
* Ratio of distance of tone hole from blow hole to tube length, L0 (L0 = 1)
Step Nine
Drill your holes 0.5 mm less as compared to your chosen values. Open just hole 1 and compare the note obtained by using a frequency meter to the note you wished to play. Sand the tone hole till you get an accurate and sharp note. If the note is a bit flat, you can enlarge the first open hole (from the blow end) a bit to sharpen it. If you drill your holes a little small to begin
२०५ with, you might be able to bring your flute into tune without having to make a second one.
Similarly, check and sharpen all the tone holes.
Now check the tuning of the flute, and estimate adjustments to make to your second flute. That is, if your note is 3% flat, move the hole, 3% closer to the blow hole, etc. A frequency counter is best for this, but if necessary, you can do it “by ear” (e.g. by comparing to another instrument which is in tune and listening to the beats). This you can do by playing different ‘Raags’ and tuning your flute for the next one month or so till you get satisfaction of all notes.
If you see systematic problems (e.g. Re is a little flat, Ga a little flatter, and by the time you get to Ni it is very flat) then your value of Les can be adjusted to fix all the holes at once. In other words, side the cork a little into the tube to tune your flute.
२०६
MATHEMATICAL CALCULATIONS FOR PANNALAL GHOSH FLUTE OF SEVEN TONE HOLES
MATHEMATICAL CALCULATIONS FOR PANNALAL GHOSH FLUTE OF SEVEN TONE HOLES
If you want to expand your traditional 6 tone hole flute into a pannalal ghosh style flute of 7 tone holes, the calculations for the first 6 holes stay the same as the program explained above. For calculating the 7 th tone hole (teevra madhyam played by little finger of right hand (for right hand flutist)), the procedure is as follows:
Considering the same safed 4 flute as in the above program, the results obtained for the first 6 holes are
Parameter Hole1 Hole2 Hole3 Hole4 Hole5 Hole6 Dthi 11 mm 11.5 mm 10 mm 11 mm 11.5 mm 10 mm 477.484 424.026 394.479 338.157 294.932 249.064 Lbti mm mm mm mm mm mm
Note: The Traditional Hindustani 6 tone hole flute has Pa as its open end note. But when the 7 th tone hole is made, Pa becomes the note playing out of the newest hole. The open end note is Ma Teevra, in case of Pannalal Ghosh flute of 7 tone holes.
Majority of the steps are same as the ones performed in the above program of traditional flute. The ones which have changed are explained below.
२०७
Step One
As an extra tone hole has to be added, the length of the flute also has to be extended (or if your original aim had been to make a 7 hole flute, then keep an extra length from the beginning). This length is calculated in a similar way as Step one of the 6 hole program.
For our example,
Frequency of Ma teevra (safed 4) = 491.1 Hz
Frequency of Kharja Teevra Ma (open end note) = 491.1 / 2 = 245.5 Hz
Tube Length = 311.88 / (2 * 245.5) = 635 mm
This is only an approximate length. For the actual distance, using a frequency meter, slowly sand this length till you get the desired note at the open end.
For our example,
The length from blow hole to open end comes to be 620 mm. Including 80 mm distance for stopper and close end, the total length Lo of the tube comes to be 620 + 80 = 700 mm
PANNALAL GHOSH – HINDUSTANI 7 TONE HOLE FLUTE
२०८
PANNALAL GHOSH – HINDUSTANI 7 TONE HOLE FLUTE Step Two
The method to calculate the position of the tone hole (Kharja Pa) is a bit different as to what we have calculated for the earlier program of traditional Hindustani flute of 6 tone holes. For our value of L0, we need to have a fictitious length L0 corresponding to the Kharja Sa of safed 4.
To get this length, we use the formula
L0 = the wavelength / 2
= (Speed of Sound / Frequency) / 2
Here, speed of sound is taken as 345 * 0.949 = 327.6
Note: 0.949 is multiplying factor for L0 open end corrections at 25 C
Frequency of Sa (safed 4) = 349.23 Hz
Frequency of Kharja Sa = 349.23 / 2 = 174.615 Hz
L0 = 327.6 / (2 * 174.615) = 938 mm
२०९
Step Three
To calculate our ratios, the note at the open end of our fictitious length (L0), Kharja Sa becomes 1.
For our example,
The 7 th tone hole we are adding is a Kharja Pa hole. Therefore, with respect to the fictitious open end Kharja Sa note, the ratio becomes 2/3
Note Number Note being played Ratio 0 Sa 1 1 Pa 2/3
Step Four
Now, the entire procedure is same as the one performed in the program of traditional Hindustani flute of 6 tone holes. Except, this is for 1 tone hole only. You need to enter your diameter of tone hole and keep the lip coverage constant, i.e. 0.1. Enter variable thickness if any. The result you get is the position of the 7th tone hole on the actual flute of length 620 mm.
For our example, the tube thickness, the tube diameter, blow hole diameter remain unchanged. The values changed are listed below
Example Variable Results L0 938 mm Dth1 9 mm
२१०
Following all other steps from step 2 to step 7 as shown in traditional flute program, we get the value of Lbt1 (distance of Pa from blow hole) = 556.272 mm
The following table summarizes all the values we have calculated
Parameter Hole1 Hole2 Hole3 Hole4 Hole5 Hole6 Hole 7
11.5 11.5 Dthi 9 mm 11 mm mm 10 mm 11 mm mm 10 mm
556.272 477.484 424.026 394.479 338.157 294.932 249.064 Lbti mm mm mm mm mm mm mm
Ratio of lengths * 0.8972 0.7701 0.6839 0.6363 0.5454 0.4757 0.4017
* Ratio of distance of tone hole from blow hole to tube length, L0 (L0 = 1)
The following table summarizes all the values we have calculated
Parameter Hole1 Hole2 Hole3 Hole4 Hole5 Hole6 Hole 7
11.5 11.5 Dthi 9 mm 11 mm mm 10 mm 11 mm mm 10 mm
556.272 477.484 424.026 394.479 338.157 294.932 249.064 Lbti mm mm mm mm mm mm mm
Ratio of lengths 0.8972 0.7701 0.6839 0.6363 0.5454 0.4757 0.4017
* Ratio of distance of tone hole from blow hole to tube length, L0 (L0 = 1)
२११
MATHEMATICAL CALCULATIONS FOR KESHAV VENU OF ELEVEN TONE HOLES
MATHEMATICAL CALCULATIONS FOR HINDUSTANI FLUTE – KESHAV VENU OF 11 TONE HOLES*
Keshav Venu is a flute with 11 holes. There are different diameters, thicknesses and lip coverages for all holes. Therefore, the calculations also have to be split up accordingly.
The program is split into 4 parts. Part 1 is for hole numbers 5 to 10 (lip coverage considered as 0.1). Part 2 is for holes 1 to 3 (lip coverage is 0). Part 3 is hole 4 (lip coverage 0.1) and Part 4 is the final hole 11, which is determined graphically.
PART 1
This is similar to determining approximate finger hole locations for a traditional 6 tone hole flute (i.e. 1 blow hole and 6 tone holes total 7 holes).
The wavelength of sound produced is determined by the flute tube and tone holes. For a pipe of length L, open at both ends, and ignoring end effects, the wavelength of sound is twice the length of the tube. The frequency produced is given by dividing the speed of sound (345 m/s) at 25 C by the wavelength.
२१२
The method described here is to estimate an effective length for a real (cylindrical) tube taking into account end effects, the size of tone holes, etc.
Let’s discuss the basic anatomy of a primitive flute, the geometries that are involved, and name and number the holes.
The basic flute is constructed of a single tube with a stopper at one end (typically cork.) This discussion will refer to this end as the “Close” end. The close end is also the end which will have the blow hole. The blow hole is the hole the instrumentalist will blow across to generate the sound (it’s the hole the player blows across.) The blow hole is located just beside the stopper.
* In illustrating examples, the Keshav Venu flute we are considering is a Safed 3 (E) Basuri of 11 tone holes with Karja Sa (Frequency 164.815 Hz) as open end note.
The other end of the flute is open and will be called the “Open” (imaginary) end for Kharja Pa.
The tone holes are the holes in the flute which the player will open or close using their fingers. The order that the holes are numbered for this discussion will be from 5 to 10, where 5 is the hole closest to the imaginary open end and 10 is the hole closest to the imaginary close end. Other conventions or definitions may apply when learning how to play such an instrument but we’re talking about this in a way to keep it as simple as possible for the following mathematical calculations. Here’s a summary table.
२१३
FLUTE TERMINOLOGIES
• Tube – The body of the flute • Stopper – The cork stopper which is closing one end of the tube. Tuning the instrument can be done by moving the stopper. • Close End – The end of the tube which has been closed by the stopper in it. • Open End – The end of the tube which is open. • Blow Hole – The hole the player will blow across to make the flute play. • Tone Hole – Holes that the user will “open” or “close” with his fingers when he plays the flute.
There are many dimensions involved with the flute. Here’s a list of geometric parameters that will be used.
२१४
VARIABLE PARAMETERS OF TRADITIONAL FLUTE
• t - wall thickness of the tube. • Lo - Length of the tube from the center of the blow hole to the open end. • Les - Effective length of the tube at the close end. • Dt - Inside diameter of the tube • Rt - Inside radius of the tube • Lbs - Distance from center of blow hole to stopper • L0 - Actual real distance from center of blow hole to open end. • Le0 - Effective distance from center of blow hole to open end. • Lbt1 - Distance between the blow hole and tone hole 1. • Lbt2 - Distance between the blow hole and tone hole 2. • Lbt3 - Distance between the blow hole and tone hole 3. • Lbt4 - Distance between the blow hole and tone hole 4. • Lbt5 - Distance between the blow hole and tone hole 5. • Lbt6 - Distance between the blow hole and tone hole 6. • Dbh - Diameter of blow hole. • Dt1 - Diameter of tone hole 1. • Dt2 - Diameter of tone hole 2. • Dt3 - Diameter of tone hole 3. • Dt4 - Diameter of tone hole 4. • Dt5 - Diameter of tone hole 5. • Dt6 - Diameter of tone hole 6. • Rbh - Radius of the blowhole.
Now, we begin with the calculations for tone holes 5 to 10 with an imaginary open end. The note at this imaginary open end is Kharja Pa.
२१५
If the fundamental (all tone holes closed) corresponds to an effective length, then the desired effective lengths for a six (or fewer) hole flute to produce notes of the Hindustani (Indian) flute are given by the following table:
Note Number Note Note actually Ratio being played 0(Imaginary) Sa Pa 1 5 Re Dha 8/9 6 Ga Ni 4/5 7 Ma(Sudha) Sa 3/ 4 8 Pa Re 2/3 9 Dha Ga 16/27 10 Ni(Teevra) Ma(Teevra) 128/243
The actual physical length of the flute will be shorter than Le0 and the distance to the tone holes (from the close end) will be shorter than the values calculated. The end effects (at both the blow hole and at the first open hole) act (approximately) as an additional length which must be subtracted from the calculated effective lengths to get the physical lengths desired. As long as you don’t deviate too much from typical flute proportions, these approximate calculations should get you pretty close.
Step One
First, make your flute with no tone holes and adjust the length to match the desired lowest note. You can do this by estimating what the tube length should be by using the following formula.
Tube Length = the wavelength / 2
= Speed of Sound / (Frequency *2)
For getting the tube length,
२१६
Frequency of Sa (E) = 329.63 Hz
Frequency of Kharja Sa (actual open end note) = 329.63 / 2 = 164.815 Hz
Tube Length = 345 / (2 * 247.22) = 1046.63 mm
This length is just an approximation. For the actual length, you will need to use a frequency meter and slowly sand off the edges till you get the desired note at the open end.
For our example, the length of the tube after sanding comes out to be 994 mm.
With reference to page no of part “Basic Principles of Flute” point number 15, the appropriate blow hole diameter for effortless playing/blowing for a Hindustani flute is 11 to 11.5 mm. This enhances the flute’s properties, such as, sound quality, tunefulness and octave range.
If the blow hole is of a bigger diameter, the tone is deeper. However, the power and stamina which needs to be exerted for playing also increases. Therefore, flutes with bigger blow hole diameters are not recommended for elaborate musical performances lasting for more than 3 hours. Also, octave range reduces to 2 or even more so.
Oval shaped blow holes are good for flute playing.
For our example,
We have taken the values of
Example Results Variable Dt 25.4 mm Dbh 12.2 mm
२१७
But for the calculations of part1, we will need to first consider a fictitious length L0.
For our example,
After sanding and corrections, L0 comes to approximately 620 mm.
Step Two
Cut some material (cork or rubber) with a diameter which will snugly fit into the tube. We now have to calculate where the “stopper” will need to be placed. In reality it doesn’t matter too much as long as the inner edge is pretty close to the blowhole.
We’ll also need to know the thickness of the tube wall. You will have to find this using calipers or use the specifications of the tube.
Example Variable Results t 2.3 mm
We need to calculate temporary value called Heff.
Heff = t + 0.85 * Dbh
For our example,
Heff = 2.3 + (0.85 * 12.2) = 12.67 mm
२१८
Now we’ll define a parameter called e where e = fraction of blow hole covered by the player’s lip and has a value typically in the range of 0 to 0.15 and in some special cases even negative. In this example, we’ll use 0.1 for a value of e. With this value now defined, it’s possible to calculate Les.
Les = Heff * (1-e) * (Dt / Dbh)^2
For our example,
Les = 12.67 * (1-0.1) * (25.4/12.2) * (25.4/12.2) = 12.67 * 0.9 * 2.174 * 2.174
= 49.427 mm
Using a single Les for all notes is a reasonable approximation IF Lbs is close to 0.37 Les. We want to know Lbs so that we know where to put the stopper that we made therefore we’ll use the following expression to calculate Lbs.
Lbs = 0.37 * Les
For our example,
Lbs = 0.37 * 49.427 = 18.288 mm
Step Three
The L0 we have calculated is
Example Variable Results L0 620 mm
We’re now ready to calculate the effective length of the tube from the center of the blowhole to the end of the tube. This is a
२१९ fictitious length and is used to compensate for end effects which exist in the real instrument. We’re going to call this effective length Le0. The first thing we need is the tube inner radius.
Rt = Dt / 2
For our example, Rt = 25.4 / 2 = 12.7 mm
Next, we can calculate Le0.
Le0 = Les + L0 + (0.6 * Rt)
For our example,
Le0 = 49.427 + 620 + (0.6 * 12.7) = 677.047 mm
Step Four
Now use your value of Le0 and the table describing the individual notes of the scale to get desired effective lengths for each of the hole positions, measured from the center of the blowhole to the center of the tone hole. Call these Lebti where ‘i’ corresponds to the tone hole number (e.g. Lebti = 8 / 9 Le0 etc.)
For our example, let’s calculate the first two.
Lebt5 = 8 / 9 * 677.047 = 601.812 mm
Lebt6 = 4 / 5 * 677.047 = 541.164 mm
The following table shows the results for all 6 values.
२२०
Example Variable Results Lebt5 601.812 mm Lebt6 541.164 mm Lebt7 507.785 mm Lebt8 451.365 mm Lebt9 401.121 mm Lebt10 356.634 mm
Let’s restructure this table little bit because we’re going to be calculating lots of numbers for each hole. As we keep on adding parameters, the table will grow.
Parameter Hole5 Hole6 Hole7 Hole8 Hole9 Hole10 601.812 541.164 507.785 451.365 401.121 356.634 Lebti mm mm mm mm mm mm
Step Five
Next, we’re going to pick our tone hole diameters and calculate thickness corrections for each hole. For our example, we’re going to use the following diameters :
Parameter Hole5 Hole6 Hole7 Hole8 Hole9 Hole10 Dthi 11 mm 13 mm 11 mm 11.5 mm 12 mm 12 mm (Note: These diameters are specifically chosen for effortless closure and the possibility of expanding the flute to have more tone holes and thus increasing the range of the flute.
For example
• Left thumb tone hole (closest hole to blow hole) for fine bridging of 2 consecutive octaves (glissando or meend). This is the soul of Hindustani classical music.
२२१
• Right little finger tone hole (fully closed) is for adding Kharja sudha Madhyam instead of teevra Madhyam of Pannalal Ghosh flute (half hole is closed for playing teevra madhyam).
The actual position of each hole will depend on the tone hole size and the position of any of any other open tone holes. Hence, you have some leeway to choose one or the other of these. Smaller tone holes will give a mellower sound and larger tone holes give a brighter (and louder) sound. Your holes do not need to be all the same size, so it is possible to make some choices which affect the ergonomics of your flute.
Below we will assume that all the tone holes are fixed in size, and only their positions are to be adjusted. If this results in tone hole positions which are uncomfortable (or unusable) then adjust one or more of the tone hole sizes and recompute.
Parameter Hole5 Hole6 Hole7 Hole8 Hole9 Hole10 Dthi 11 mm 13 mm 11 mm 11.5 mm 12 mm 12 mm 601.812 541.164 507.785 451.365 401.121 356.634 Lebti mm mm mm mm mm mm
This provides enough information to calculate the thickness correction lengths.
Ltci = t * (Dthi/Dt)^2 / 4 + (All Ltcl’s)
Where: j = the number of closed holes between the first open hole and the blow end
२२२
The expression is attempting to show the correction factor for a specific hole ‘i’ and includes all the thickness corrections for all holes between hole i and the blow hole. The value of j represents the number of holes which are included in the thickness correction.
२२३
Parameter Hole5 Hole6 Hole7 Hole8 Hole9 Hole10 11.5 Dthi 11 mm 13 mm 11 mm mm 12 mm 12 mm
601.812 541.164 507.785 451.365 401.121 356.634 Lebti mm mm mm mm mm mm
Value of i 5 6 7 8 9 10
Value of j 9 8 7 6 5 4
For our example, we will calculate all of them because this is a confusing parameter. We have to work our way from hole 10 to hole 5. This is because we have to sum all the thickness corrections for each of the tone holes as we work our way from the blow hole to the open end of the flute.
Ltc10 = 0 (There are no holes in between hole number 10 and the blowhole, therefore, no correction is required)
Ltc9 = 2 * (12 / 25.4) * (12 / 25.4) / 4 + (0) = 0.12834 mm
Ltc8 = 2 * (11.5 / 25.4) * (11.5 / 25.4) / 4 + (0.12834) = 0.24621 mm
Ltc7 = 2 * (11 / 25.4) * (11 / 25.4) / 4 + (0.24621) = 0.35401 mm
Ltc6 = 2 * (13 / 25.4) * (13 / 25.4) / 4 + (0.35401) = 0.5047 mm
Ltc5 = 2 * (11 / 25.4) * (11 / 25.4) / 4 + (0.5047) = 0.6125 mm
२२४
The following table shows the thickness corrections for each of the holes.
Parameter Hole5 Hole6 Hole7 Hole8 Hole9 Hole10 Dthi 11 mm 13 mm 11 mm 11.5 mm 12 mm 12 mm
601.812 541.164 507.785 451.365 401.121 356.634 Lebti mm mm mm mm mm mm
Value of i 5 6 7 8 9 10
Value of j 9 8 7 6 5 4
0.6125 0.5047 0.35401 0.24621 0.12834 Ltci mm mm mm mm mm 0
Ultimately, we’re also going to need the thickness correction factor when all the tone holes are open. This will be the thickness correction for the entire tube length and corresponds to a value of j = 10.
For our example,
Ltc0 = 2 * (12/25.4) * (20/25.4) /4 + 0.6125 = 0.7408 mm
Step Six
We’re now going to make a finer approximation for L0 and then working our way from tone hole 5 to tone hole 10. The first step is to compute the approximation.
L0f = L0 – Ltc0
२२५
For our example,
L0f = 620 – 0.7408 = 619.26 mm
Step Seven
The next is to compute the length for hole number 5. Computations of hole length require iteration. This means that the computation will need to be repeated several times substituting some results from a previous step into the new step until the results stop changing. This may seem complicated at first but the example will show you how to do it.
We need a jump off point so we’re going to calculate the initial length using the following formula.
L0bti = Lebti – Les – (0.6 * Rt)
Note: L0bti : Blow hole to tone hole distance for hole number ‘i’
Lebti : Effective blow hole to tone hole distance for hole number ‘i’
The expression takes into account the correction for the length at the close end and the value of 0.6 * Rt is the length correction at the close end.
For our example,
२२६
L0bt5 = 601.812 – 49.427 – (0.6 * 12.7) = 544.765 mm
The first hole uses the following formulas.
K1 = 0.75 * Dth5 + t
K2 = (Dth5 / Dt) ^2
D = L0f – Lbt5
Lc5 = K1 / (K2 + (K1 / D))
For our example, we’ll calculate K1 and K2 which don’t change throughout the iterative process.
K1 = 0.75 * 11 + 2.3 = 10.55 mm K2 = (11 / 25.4) * (11 / 25.4) = 0.1875 Parameter Hole5 Hole6 Hole7 Hole8 Hole9 Hole10 11.5 Dthi 11 mm 13 mm 11 mm mm 12 mm 12 mm
601.812 541.164 507.785 451.365 401.121 356.634 Lebti mm mm mm mm mm mm
Value of i 5 6 7 8 9 10
Value of j 9 8 7 6 5 4
0.6125 0.5047 0.3541 0.2462 0.1283 Ltci mm mm mm mm mm 0
10.55 K1 mm
K2 0.1875
२२७
Now we start the iteration process. The iterative formula is as follows.
Lbti(new) = Lebti – Les – Lci - Ltci
We’ll start the first iteration for hole 5.
D = 619.26 - 544.765 = 74.495 mm
Lc5 = 10.55 / (0.1875 + (10.55 / 74.495)) = 32.0551 mm
Lbt5(new) = 601.812 - 49.427 – 32.0551 – 0.4852
= 518.8447 mm
Second Iteration; use the last version of Lbt5(new) to recompute D. Proceed with the iteration.
D = 619.26 - 518.8447 = 100.4153 mm
Lc5 = 10.55 / (0.1875 + (10.55 / 100.4153)) = 36.061 mm
Lbt5(new) = 601.812 - 49.427 – 36.061 – 0.4852 = 515.8388 mm
Continue the process until Lbt5(new) stops changing. The final result will be Lbt5 which is the actual length which should be used to drill hole 5.
The final result after 6 iterations is 515.276 mm.
२२८
Step Eight
The calculation for the next 5 holes will use a different set of formulas than for hole number 5 but the steps are very similar. Here are the formulas.
K1 = 0.75 * Dthi + t
K2 = (Dt / Dthi) ^2
Note: K2 for the tone holes other than 1 has an inverted ratio than hole 5.
S = (Lbt(i-1) – Lbti) / 2
Lc6 = S * [(1 + 2 * K1 * K2 / S) ^1/2 - 1]
Let’s work on hole number 2. We’ll calculate K1 and K2 which don’t change throughout the iterative process.
K1 = 0.75 * 13 + 2.3 = 12.05 mm
K2 = (25.4 / 13) * (25.4 / 13) = 3.818
Because the tone hole diameters are all different for our example problem, K1 and K2 are different for each of the holes 6 to 10.
२२९
Parameter Hole5 Hole6 Hole7 Hole8 Hole9 Hole10 Dthi 11 mm 13 mm 11 mm 11.5 mm 12 mm 12 mm
601.812 541.164 507.785 451.365 401.121 356.634 Lebti mm mm mm mm mm mm
Value of i 5 6 7 8 9 10
Value of j 9 8 7 6 5 4
0.6125 0.5047 0.3541 0.2462 0.1283 Ltci mm mm mm mm mm 0
10.55 12.05 10.55 10.925 11.3 K1 mm mm mm mm mm 11.3 mm
K2 0.1875 3.818 5.332 4.878 4.48 4.48
515.276 Lbti mm
We need the jump off point to start the iteration process just like we did for step seven.
For our example,
L0bt6 = 541.164 - 49.427 – (0.6 * 12.7) = 484.117 mm
Now, we’ll start the iterative process using the same iterative formula as in step seven.
We’ll start the first iteration for hole 6.
S = (515.276–484.117) / 2 = 15.343 mm
२३०
Lc6 = 15.343 * [(1 + 2 * 12.05 * 3.818 / 15.343) ^1/2 - 1]
= 25.242 mm
Lbt6(new) = 541.164 - 49.427 - 25.242 – 0.5047
= 465.99 mm
Second iteration; use the last version of to recompute S. Proceed with the iteration.
S = (515.276 – 465.99) / 2 = 24.643 mm
Lc6 = 24.643 * [(1 + 2 * 12.05 * 3.818 / 24.643) ^1/2 - 1]
= 28.974 mm
Lbt6(new) = 541.164 - 49.427 - 28.974 – 0.5047
= 462.258 mm
Continue the process until Lbt6(new) stops changing. The final result will be Lbt6 which is the actual length which should be used to drill hole 6.
The final result after 6 iterations is 462.149 mm.
Repeating the same iterative procedure for holes 7 to 10 will provide the final hole positions.
२३१
Parameter Hole5 Hole6 Hole7 Hole8 Hole9 Hole10 11.5 Dthi 11 mm 13 mm 11 mm mm 12 mm 12 mm
601.812 541.164 507.785 451.365 401.121 356.634 Lebti mm mm mm mm mm mm
Value of i 5 6 7 8 9 10
Value of j 9 8 7 6 5 4
0.6125 0.5047 0.3541 0.2462 0.1283 Ltci mm mm mm mm mm 0
10.55 12.05 10.55 10.925 11.3 11.3 K1 mm mm mm mm mm mm
K2 0.1875 3.818 5.332 4.878 4.48 4.48
515.276 462.149 428.078 367.607 321.106 277.143 Lbti mm mm mm mm mm mm
Step Nine
Drill your holes 0.5 mm less as compared to your chosen values. Open just hole 5 and compare the note obtained by using a frequency meter to the note you wished to play. Sand the tone hole till you get an accurate and sharp note. If the note is a bit flat, you can enlarge the first open hole (from the blow end) a bit to sharpen it. If you drill your holes a little small to begin with, you might be able to bring your flute into tune without having to make a second one.
Similarly, check and sharpen all the tone holes.
२३२
Now check the tuning of the flute, and estimate adjustments to make to your second flute. That is, if your note is 3% flat, move the hole, 3% closer to the blow hole, etc. A frequency counter is best for this, but if necessary, you can do it “by ear” (e.g. by comparing to another instrument which is in tune and listening to the beats)
If you see systematic problems (e.g. Re is a little flat, Ga a little flatter, and by the time you get to Ni it is very flat) then your value of Les can be adjusted to fix all the holes at once. In other words, side the cork a little into the tube to tune your flute.
PART 2
Majority of the steps are same as the ones performed in the above program. The steps which have changed are explained below.
Step One
We are now calculating for the lower 3 holes (hole numbers 1, 2, 3). The value of L0 we take here is the actual tube length which we have cut and sanded to match our desired note, Kharja Sa (Frequency 164.815 Hz).
For our example,
L0 = 994.3 mm
Step Two
The ratios are calculated by taking ratio of Kharja Sa as 1. The other ratios are taken correspondingly.
२३३
Note Number Note being played Ratio 0 Sa 1 1 Re 8/9 2 Ga 4/5 3 Ma (Sudha) 3/ 4
Note: The ratio of Ma here is 3/ 4 as opposed to the ratio of Ma taken in part 1. This is because the Ma taken in this part is Sudha, whereas the Ma in part 1 was teevra.
Step Three
Here, because the notes are kharja and being feeble, the lip coverage is no longer 0.1. It changes to 0, to make the note clear and louder.
Therefore, for this part, e = 0.
Step Four
We have considered a different thickness other than the pipe diameters for these holes. This has been done to provide keys for playing notes with right hand thumb.
Example Variable Results t1 6 mm t2 6 mm t3 6.5 mm
For the hole which has a different thickness than pipe diameter, the calculations change a bit. In all the formulas of the length procedure shown in part 1, all the thicknesses (t) are to be
२३४ replaced by individual thicknesses (t1/t2/t3 depending on the hole).
Step Five
Now, the entire procedure is same as the one performed in the above part 1. You need to enter your diameter of tone hole and keep the lip coverage constant. The result you get is the position of the lower 3 tone holes on the actual flute of length 994.3 mm
For our example, the tube thickness, the tube diameter, blow hole diameter remain unchanged. The values changed are listed below
Example Variable Results Dth1 14 mm Dth2 10 mm Dth3 8.5 mm
The following table summarizes all the values we have calculated
Parameter Hole1 Hole2 Hole3 Dthi 14 mm 10 mm 8.5 mm Ti 6 mm 6 mm 6.5 mm 843.967 732.905 683.566 Lbti mm mm mm
Note: These three tone holes are to be drilled on the center line diametrically opposite to the blow hole center line and about 2 mm (10 degrees) towards the player. This offers better support to the hand and extra upward force while playing keys.
२३५
PART 3
This part calculates the single hole left in between, tone hole number 4. (Kharja Pa; which was the imaginary open end note in part 1).
The procedure is exactly same as part 2, regarding extra thicknesses, ratios and L0. The only different factor is that lip coverage is no longer 0, but is 0.1, same as part 1, i.e., e = 0.1
These are the values we have taken for calculating the distance of Pa
Example Variable Results Dth4 10 mm t4 7.5 mm
And the ratios taken are
Note Number Note being played Ratio 0 Sa 1 4 Pa 2/3
Calculating, the value we get for distance of tone hole 4 (Kharja Pa) from blow hole is 572.776 mm.
Note: This hole is to be drilled at right angles to the center line of the first 6 holes; towards the player, about 2 mm (10 degrees) above the center line.
२३६
PART 4
11th hole (of left hand thumb for right hand flutist), Pa cannot be calculated by the method we have been using till now. It has to be graphically calculated as follows:
Calculate the distance Lo between tone hole 10 and tone hole 9.
For our example,
Lbt10 = Lbt10 from part1 = 277.143 mm
Lbt9 = Lbt9 from part1 = 321.106 mm
Distance between tone hole 10 and 9 = 321.106 – 277.143 = 43.963 mm
Lo = 43.963 / 2 = 21.9815 mm
This distance L0 is the distance between tone hole 11 and 10.
Distance Lbt11 (Distance between tone hole 11 and the blow hole) = L0bt10 – Lo
For our example,
Lbt 11 = 277.143 – 21.9815 = 255.1615 mm
Diameter of this hole should be kept between 6 to 8 mm and is to be placed on the curvature of the tube as per the convenience of flutist.
For our example, we have taken a diameter of 8mm
The diameter should be initially drilled to 5 mm and then with the help of a frequency meter slowly sanded and deburred till the required note is achieved.
२३७
Note: This hole is to be drilled at right angles to the center line of the first 6 holes; away from the player, about 2 mm (10 degrees) below the center line.
The following tables summarize all the values we have calculated
Parameter Hole1 Hole2 Hole3 Hole4 Hole5 Hole6
Dthi 14 mm 10 mm 8.5 mm 9 mm 11 mm 13 mm
Ti 6 mm 6 mm 6.5 mm PT* PT* PT*
843.967 732.905 683.566 572.776 515.276 462.149 Lbti mm mm mm mm mm mm
Ratio of Lengths* 0.8488 0.7371 0.6422 0.576 0.5182 0.4648
Parameter Hole7 Hole8 Hole9 Hole10 Hole11
Dthi 11 mm 11.5 mm 12 mm 12 mm 8 mm
Ti PT* PT* PT* PT* PT*
428.078 367.607 321.106 277.143 255.162 Lbti mm mm mm mm mm
Ratio of Lengths* 0.4305 0.3697 0.3229 0.2787 0.2566
* Ratio of distance of tone hole from blow hole to tube length, L0 (L0 = 1)
* PT = Pipe thickness
२३८
Traditional Hindustani 6 tone hole Flute
Pannalal Ghosh 7 tone hole Flute
Keshav Venu 11 tone hole Flute
२३९
MATHEMATICAL CALCULATIONS FOR MURCHHANA FLUTE FOR RAAG TODI KOMAL RISABH AS SHADJA - SAFED 3 (E) I.E. KALI 2 (E FLAT) FLUTE
MATHEMATICAL CALCULATIONS FOR MURCHHANA FLUTE FOR RAAG TODI -KOMAL RISABH AS SHADJA SAFED 3 (E) I.E. KALI 2 (E FLAT) FLUTE*
The calculations for the distances between blow hole and tone holes for this flute is same as the procedure we followed for Keshav Venu, except it is only for 8 tone holes, while Keshav Venu was for 11 tone holes.
Accordingly, the program is split into 2 parts. Part 1 is for tone hole numbers 3 to 8 (lip coverage considered as 0.1). Part 2 is for tone holes 1 and 2 (lip coverage is 0).
२४०
PART 1
This is same as Part 1 of Keshav Venu program.
For our calculations of the tone holes 3 to 8, we need to consider an imaginary open end. The note at this imaginary open end is Kharja Pa.
Therefore, the ratios of the other tone holes correspondingly become:
Note Number Note Note actually Ratio being played 0(Imaginary) Sa Pa 1 3 Re Dha 8/9 4 Ga Ni 4/5 5 Ma(Sudha) Sa 3/ 4 6 Pa Re 2/3 7 Dha Ga 16/27 8 Ni(Teevra) Ma(Teevra) 128/243
Step One
We need to calculate our actual tube length. For this, we use the formula
Tube Length = the wavelength / 2
२४१
= (Speed of Sound / Frequency) / 2
For our example, the open end note is Kharja Sudha Ga (Frequency = 185.835 Hz)
Therefore, Tube Length = 345 / (2 * 185.835) = 928 mm
Adding 80 mm for the stopper and adjustment of the blow hole, we get the initial tube length to 928 + 80 = 1028 mm, rounded to 1030 mm. Sanding out till we get the exact note, the blow hole to open end distance for our flute comes out to be 807 mm.
But for the calculations of part 1, we will need to first consider a fictitious length L0 corresponding to Kharja Pa of Kali 2.
For our example,
After sanding and corrections, L0 comes to approximately 640 mm.
Step Two
Following all other steps as in Keshav Venu Part 1, we get the following parameters:
Variable Example Results L0 640 mm Dt 26.5 mm Dbh 13 mm t 2.5 mm e 0.1 lip coverage
२४२
Variable Example Results Dth3 11 mm Dth4 13.5 mm Dth5 12.5 mm Dth6 11.5 mm Dth7 12.5 mm Dth8 12 mm
Step Three
Following the same as for the calculations as Keshav Venu Part 1, we get the values:
Parameter Hole3 Hole4 Hole5 Hole6 Hole7 Hole8 13.5 12.5 11.5 Dthi 11 mm mm mm mm 12.5 mm 12 mm 529.947 476.881 444.484 376.966 331.711 284.275 Lbti mm mm mm mm mm mm
PART 2
All the steps to be calculated for this part are same as for Keshav Venu part 2, except that this is only for 2 tone holes, while Keshav Venu Part 2 was for 3 tone holes.
Step One
Calculating a fictitious length corresponding to the Kharja Sudha Ga of Kali 2,
For our example,
L0 = 807 mm
Step Two
The Ratios taken are
२४३
Note Number Note being played Ratio 0 Ga 40/49 1 Ma 3/ 4 2 Pa 2/3
Note: The Ga ratio is taken as 1/1.225 = 40/49 as sudha ga, ati komal ga, komal ga are also played at same hole by varying air pressure and lip coverage between 0 and 0.15.
But, for our calculations, we need to have the open end or number 0 note (Ga) as 1. Therefore, multiplying all other ratios by 49/40 we get
Note Number Note being played Ratio 0 Ga 1 1 Ma 147/160 2 Pa 98/120
Step Three
Following all other steps same as Keshav Venu Part 2, we get the following parameters:
Therefore, for this part, e = 0.
Example Variable Results t2 5 mm / PT* Dth1 11 mm Dth2 9 mm *PT = Pipe Thickness
The tube inside diameter, pipe thickness and blow hole diameter remain same as Part 1.
२४४
The following table summarizes all the values we have calculated
Parameter Hole1 Hole2 Dthi 11 mm 9 mm 5 mm / Ti PT* PT* 590.109 mm/ 705.442 602.23 Lbti mm mm *PT = Pipe Thickness
Note: These two tone holes are to be drilled on the center line diametrically opposite to the blow hole center line and about 2 mm (10 degrees) towards the player. This offers better support to the hand and extra upward force while playing keys.
The final results calculated are:
Parameter Hole1 Hole2 Hole3 Hole4 Hole5 Hole6 13.5 12.5 11.5 Dthi 11 mm 9 mm 11 mm mm mm mm Ti PT* 5 mm PT* PT* PT* PT* 590.109 mm/ 705.442 602.23 529.947 476.881 444.484 376.966 Lbti mm mm mm mm mm mm Ratio of 0.733 / Lengths* 0.876 0.746 0.658 0.592 0.552 0.468
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Parameter Hole7 Hole8 Dthi 12.5 mm 12 mm Ti PT* PT* 331.711 284.275 Lbti mm mm Ratio of Lengths* 0.412 0.353
* Ratio of distance of tone hole from blow hole to tube length, L0 (L0 = 1)
*PT = Pipe thickness
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MATHEMATICAL CALULATIONS FOR MURCHHANA FLUTE FOR RAAG BHAIRAVI SUDHA GANDHAR AS SHADJA - SAFED 3 (E) I.E. SAFED 1 (C) FLUTE
MATHEMATICAL CALULATIONS FOR MURCHHANA FLUTE FOR RAAG BHAIRAVI – SUDHA GANDHAR AS SHADJA SAFED 3 (E) I.E. SAFED 1 (C) FLUTE
The calculations for the distances between blow hole and tone holes for this flute is same as the procedure we followed for Keshav Venu, except it is only for 8 tone holes, while Keshav Venu was for 11 tone holes.
Accordingly, the program is split into 2 parts. Part 1 is for tone hole numbers 3 to 8 (lip coverage considered as 0.1). Part 2 is for tone holes 1 and 2 (lip coverage is 0).
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PART 1
This is same as Part 1 of Keshav Venu program.
For our calculations of the tone holes 3 to 8, we need to consider an imaginary open end. The note at this imaginary open end is Kharja Pa.
Therefore, the ratios of the other tone holes correspondingly become:
Note Number Note Note actually Ratio being played 0(Imaginary) Sa Pa 1 3 Re Dha 8/9 4 Ga Ni 4/5 5 Ma(Sudha) Sa 3/ 4 6 Pa Re 2/3 7 Dha Ga 16/27
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8 Ni(Teevra) Ma(Teevra) 128/243
Step One
We need to calculate our actual tube length. For this, we use the formula
Tube Length = the wavelength / 2
= (Speed of Sound / Frequency) / 2
For our example, the open end note is Kharja Sudha Ga (Frequency = 165.56 Hz)
Therefore, Tube Length = 345 / (2 * 165.56) = 1042 mm
Adding 80 mm for the stopper and adjustment of the blow hole, we get the initial tube length to 1042 + 80 = 1122 mm, rounded to 1120 mm. Sanding out till we get the exact note, the blow hole to open end distance for our flute comes out to be 942 mm.
But for the calculations of part1, we will need to first consider a fictitious length L0 corresponding to Kharja Pa of Safed 1.
For our example,
After sanding and corrections, L0 comes to approximately 750 mm.
Step Two
Following all other steps as in Keshav Venu Part 1, we get the following parameters:
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Variable Example Results L0 750 mm Dt 29 mm Dbh 13 mm t 2.5 mm e 0.1 lip coverage
Variable Example Results Dth3 10 mm Dth4 12.5 mm Dth5 11.5 mm Dth6 10 mm Dth7 11 mm Dth8 12 mm
Step Three
Following the same for the calculations as Keshav Venu Part 1, we get the values:
Parameter Hole3 Hole4 Hole5 Hole6 Hole7 Hole8 12.5 11.5 Dthi 10 mm mm mm 10 mm 11 mm 12 mm 615.55 555.653 517.294 433.481 383.858 332.511 Lbti mm mm mm mm mm mm
PART 2
All the steps to be calculated for this part are same as for Keshav Venu part 2, except that this is only for 2 tone holes, while Keshav Venu Part 2 was for 3 tone holes.
२५०
Step One
Calculating a fictitious length corresponding to the Kharja Sa of Safed 1,
For our example,
L0 = 1252.2 mm
Step Two
The ratios are calculated by taking ratio of Kharja Sa as 1. The other ratios are taken correspondingly.
Note Number Note being played Ratio 0 Sa 1 1 Ma 3/ 4 2 Pa 2/3
Step Three
Following all other steps same as Keshav Venu Part 2, we get the following parameters:
Therefore, for this part, e = 0.
Example Variable Results Dth1 10 mm Dth2 9 mm
The tube inside diameter, pipe thickness and blow hole diameter remain same as Part 1.
२५१
The following table summarizes all the values we have calculated
Parameter Hole1 Hole2 Dthi 10 mm 9 mm 859.481 757.84 Lbti mm mm Note: These two tone holes are to be drilled on the center line diametrically opposite to the blow hole center line and about 2 mm (10 degrees) towards the player. This offers better support to the hand and extra upward force while playing keys.
The final results calculated are:
Parameter Hole1 Hole2 Hole3 Hole4 Hole5 Hole6 Dthi 10 mm 9 mm 10 mm 12.5 mm 11.5 mm 10 mm 859.481 757.84 615.55 555.653 517.294 433.481 Lbti mm mm mm mm mm mm Ratio of Lengths* 0.912 0.804 0.653 0.590 0.549 0.460
Parameter Hole7 Hole8 Dthi 11 mm 12 mm 383.858 332.511 Lbti mm mm Ratio of Lengths* 0.407 0.353
* Ratio of distance of tone hole from blow hole to tube length,
L0 (L0 = 1)
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