differential-geometrymathematical physicsparametrization
Has anyone worked out a physically justified equation
(perhaps parametrized) for the characteristic
(2D outline) shape of a candle flame?
Just one half suffices, as it is clearly symmetric about a vertical
(in the absence of wind).
(Image from this link.)
Motivation: To make realistic candle flames in computer graphics.
We recognize the shape when we see it. And when it is not quite right,
we notice the discrepancy.
I want to get it right.
Yes since {E2, E3} is an orthonormal basis of the tangent plane, $v^2 E_2 + v^3 E_3 = v$
It is our own physics assumption in force modeling.
There are no concentrated forces kept at the vertex or at any other point on the cable. All the weight $ W=\int q ds $ that acts comes from the the cable uniformly distributed along its length.
To find the shape of the cable we make a calculation on the basis of existence of a constant minimum tension $H \text{ = your} \;T_0$ at cable bottom point of horizontal tangency at $V$.
An assumption is made that the tension $T$ increases from minimum $H $ and increases as the arc length from bottom point V can be represented by three forces of a right triangle.
We assumed from physics of static equilibrium that
$$ T^2= H^2+ W_{cable}^2 \text { or} \quad T^2= H^2 +{\int q ds} ^2$$
That is, we made an a priori underlying starting physics assumption that at point $V$ it is given:
$$T=H$$
and so far it is not a consequence of any mathematical calculation.
The calculation is carried out thus:
$$ \tan \phi =\dfrac{\int q ds}{H} ;\text{ Let} \dfrac{H}{q}= c \tag 1$$
Differentiate w.r.t $x$ on which it is primed,
$$ \sec ^2 \phi \dfrac{d\phi}{ds}\cdot \dfrac{ds}{dx}=\dfrac{\sec \phi}{c}, \tag2 $$
$$\text{because} \cos \phi= \dfrac{dy}{ds} \text{ and } \dfrac{d\phi} {ds}= \dfrac{y''}{({1+y^{'2}})^{3/2}} ,\tag3 $$
it leads to the ode of catenary
$$\dfrac{y^{''}}{\sqrt{1+y^{'2}}}=\dfrac{1}{c}\tag4 $$
Let $$y'= \sinh \dfrac{x}{c} \tag5 $$
RHS of (4) then reduces to $1/c$ okay.
Integrating the above with IC $ x=0, y=c$ we can define shape of catenary:
$$ y=c \cosh\frac{x}{c} \tag6 $$
At any point including V ( where it seems indeterminate but is actually a constant)
$$H= \dfrac{W }{\tan \phi}. \tag7$$
Best Answer
I Googled "Heat equation applied to a burning candle" and found
Analysis of Burning Candle