- #1
Chrono G. Xay
- 91
- 3
I have been wanting to write an equation which predicts the tension on a circular membrane (AKA drumhead). However, I'm not sure that my answer is on the right kind of track, if it's even correct.
As for the procedures I took, first I started with the equation for tension on a string, which I obtained from the website of guitar string company D'Addario:
T = UW ( 2 L f )^2 ,
where
'UW' is the unit weight of string per unit of linear measure,
'L' is the speaking length of string, and
'f' is the fundamental frequency produced by that length of string when set in motion.
Next, I defined 'UW' as
UW = V ρ ,
where
'V' is the volume of the string and
'ρ' is the string's material density.
Then, I defined 'V' as
V = A l ,
where
'A' is the cross-sectional area of the string and
'l' is the unit length of measure (a variable which apparently doesn't contribute to the unit of measure of the equation at all).
Now *that* was a string, but if you were to take that string and stretch it over a drum so that it intersects the drum's center, could you not abstractly say then that the string now represents a diametric force line that the drumhead naturally experiences when under tension?
For a drumhead we go back the part where we define the geometry of the body under tension, 'V', but instead of the very long, thin cylinder of a string before we now have a short, very large cylinder.
V = A h π ( D / 2 )^2
where
'A' is the square unit of measure (which, like with the string, does not contribute to the equation's overall unit of measure), and
'h' is the height (or perhaps thickness) of the drumhead.
Substituting this back into the equation, and instead of 'L', being the speaking length of string, writing 'D', the speaking diameter of the drumhead, we arrive at
T = h π ( D / 2 )^2 ρ ( 2 D f )^2
Simplifying...
T = π h ρ f^2 D^4
but this doesn't have the right measure of units yet- there's one extra power of the diameter than we need to have our measure in Newtons, so there's something else we have to do: we need to divide everything by half of the drum's circumference, π ( D / 2 ) . This ends up with
T = 2 h ρ f^2 D^3
For ease of use by the layman musician, we can furthermore divide by the force of gravity to get a value in either pounds or kilograms.
As for the procedures I took, first I started with the equation for tension on a string, which I obtained from the website of guitar string company D'Addario:
T = UW ( 2 L f )^2 ,
where
'UW' is the unit weight of string per unit of linear measure,
'L' is the speaking length of string, and
'f' is the fundamental frequency produced by that length of string when set in motion.
Next, I defined 'UW' as
UW = V ρ ,
where
'V' is the volume of the string and
'ρ' is the string's material density.
Then, I defined 'V' as
V = A l ,
where
'A' is the cross-sectional area of the string and
'l' is the unit length of measure (a variable which apparently doesn't contribute to the unit of measure of the equation at all).
Now *that* was a string, but if you were to take that string and stretch it over a drum so that it intersects the drum's center, could you not abstractly say then that the string now represents a diametric force line that the drumhead naturally experiences when under tension?
For a drumhead we go back the part where we define the geometry of the body under tension, 'V', but instead of the very long, thin cylinder of a string before we now have a short, very large cylinder.
V = A h π ( D / 2 )^2
where
'A' is the square unit of measure (which, like with the string, does not contribute to the equation's overall unit of measure), and
'h' is the height (or perhaps thickness) of the drumhead.
Substituting this back into the equation, and instead of 'L', being the speaking length of string, writing 'D', the speaking diameter of the drumhead, we arrive at
T = h π ( D / 2 )^2 ρ ( 2 D f )^2
Simplifying...
T = π h ρ f^2 D^4
but this doesn't have the right measure of units yet- there's one extra power of the diameter than we need to have our measure in Newtons, so there's something else we have to do: we need to divide everything by half of the drum's circumference, π ( D / 2 ) . This ends up with
T = 2 h ρ f^2 D^3
For ease of use by the layman musician, we can furthermore divide by the force of gravity to get a value in either pounds or kilograms.
