Drum tuning bible

Kick Drum Tuning• It looks cool.• They do not like the feel of the beater on the batter head surface, it bounces as a result of notenough air relief.• They need to mic the drum from or capture the sound from the inside.• They want more projection without using a mic (less bass impact, more beater attackpresence).For those who want it because it looks cool, there is an acoustic impact on the sound by placing a hole orholes in the resonant side. By acoustic impact I mean that the removal of head material does affect thebass portion of the note coming from the drum.Allot of the “bass” portion of what you hear is based upon the surface area in the center of the drum. Thatsurface area is a diaphragm working much like a speaker radiator might work, in that it will aid in movingair. Remember that pitch is dictated by the tension and the surface area in movement. So if you remove alarge center portion, you lose a large portion of the bass reinforcement that gets emitted by the headsmovement and tension usually has to increase to compensate for the removal of the center area. Addingholes does not increase bass content as might be the case on a tuned vented speaker cabinet would.Thicker heads tend to stay in motion longer. With loose tension they will vibrate at a lesser rate, which alltranslates into lower pitch and a longer resonance. This assumes no internal muffling, or other devices tomake the head stop its vibrating motion sooner.Some want the different feel created by having air relief but still want maximum bass affect. As youremove more head area you trade off deep bass for a different feel. A solution is using smaller holesplaced around the perimeter of the head. If you want the mic to capture sound from inside, you eitherhave to resort to say the May mic system or revert to a larger hole to get the mic into the drum as yourequire. What you ultimately do will be based upon the balcance of all the factors that are important to you.It is the area of the hole that counts. Where it is located matters little for the affect on sound (as long as itisn't on the batter side). If you want maximum tone out of the head, then the size of each hole needs to bein the 1-2” size, and they need be placed closer to the perimeter, but not placed so the edge of the hole iscloser than about 1” to the break for the bearing edge. In other words, for the best tone, you need to keepas much of the center of the resonant head intact as possible. And again, it’s not the number, it’s the areadisplaced that can make a big difference and where that area is removed. You can make any number youwant, in the following example to illustrate the concept we'll make two holes to represent the maximumarea displaced by a common 7 inch hole.The math is simple. We first need to calculate the area of a 7” hole. To do this we use the formula Pi(R²). So first find half of the diameter of the 7" hole (the radius), which is 3.5”. Now multiply thattimes itself. So 3.5 x 3.5 = 12.25. Then take this result of 12.25 and multiply it times Pi, which is3.142. So we now have 12.25 x 3.142 = 38.5. So the area of the 7 inch hole we started with is 38.5square inches. This 38.5 sq. in. is important. We will simply round it up to 40 square inches, causeclose is enough.Now we can use any number of holes as long as is does not cumulatively exceed 40 square inchesof total area. Yet at the same time does equal 40 square inches. This will be the same air relief ashaving one 7” hole and the end result will be more center surface are and a stronger bass affect.Now take the 40 sq. in. and divide by 2, 3, or 4, what ever. Let’s say you want 3 holes. 40 ÷ 3 =13.33. So 13.33 is the maximum area for each of the 3 holes. So we now take the 13.33 ÷ Pi(which is 3.142) = 4.24. Now extract the square root (from a math table or calculator) of 4.24 andyou get 2.06. So 2 x 2.06 = 4.12. This means 3 holes of 4.12 diameter will give the same acousticresult as a single 7” hole.Let’s say you have1 hole of a diameter of 4.5”, a common bass drum hole. Let’s compute the areadisplaced by that single 4.5” hole. (Math: 4.5 ÷ 2 = 2.25, THEN 2.25 x 2.25 = 5.0625, THEN 5.0625x 3.142 = 15.9). A 4.5” hole has an area of 15.9 sq. in.In the above example we show that if we were to use 2 holes of 4.5”, the cumulative affect will haveless area (31.8 sq. in. total) than that of a single 7” hole, whichj we learned was about 40 sq. in. The2 – 4.5” holes will therefore be a little more bass heavy than will a head with a 7” hole because theydo not remove as much of the heads surface, although you probably will not hear it.http://home.earthlink.net/~prof.sound/id6.html (2 of 4)1/22/2005 4:11:45 PM