An ellipse, whose semi-axes have length $a$ and $b$, rolls without slipping on the curve $y=c\sin{\left(\frac{x}{a}\right)}$. How are $a,b,c$ related, given that the ellipse completes one revolution when it traverses one period of the curve?
To solve the problem, we need to establish the relationship between the semi-axes \(a\) and \(b\) of the ellipse and the amplitude \(c\) of the sine curve \(y = c \sin\left(\frac{x}{a}\right)\). The key condition is that the ellipse completes one revolution when it traverses one period of the sine curve.

1. **Period of the Sine Curve:**
   The period of the sine curve \(y = c \sin\left(\frac{x}{a}\right)\) is given by:
   \[
   T = 2\pi a
   \]
   This is because the argument of the sine function, \(\frac{x}{a}\), completes one full cycle (from \(0\) to \(2\pi\)) when \(x\) ranges from \(0\) to \(2\pi a\).

2. **Circumference of the Ellipse:**
   The circumference \(C\) of an ellipse with semi-major axis \(b\) and semi-minor axis \(a\) is approximately given by Ramanujan's approximation:
   \[
   C \approx \pi \left[ 3(a + b) - \sqrt{(3a + b)(a + 3b)} \right]
   \]
   However, for simplicity, we can use the approximation:
   \[
   C \approx 2\pi \sqrt{\frac{a^2 + b^2}{2}}
   \]

3. **Rolling Without Slipping Condition:**
   For the ellipse to roll without slipping and complete one revolution over one period of the sine curve, the circumference of the ellipse must equal the period of the sine curve:
   \[
   2\pi \sqrt{\frac{a^2 + b^2}{2}} = 2\pi a
   \]

4. **Solving for \(b\):**
   Simplifying the above equation:
   \[
   \sqrt{\frac{a^2 + b^2}{2}} = a
   \]
   Squaring both sides:
   \[
   \frac{a^2 + b^2}{2} = a^2
   \]
   Multiplying both sides by 2:
   \[
   a^2 + b^2 = 2a^2
   \]
   Subtracting \(a^2\) from both sides:
   \[
   b^2 = a^2
   \]
   Taking the square root of both sides:
   \[
   b = a
   \]

5. **Relating \(c\) to \(a\) and \(b\):**
   Given that \(b = a\), we need to consider the curvature of the sine curve and the ellipse. The maximum curvature of the sine curve \(y = c \sin\left(\frac{x}{a}\right)\) is:
   \[
   \kappa_{\text{sine}} = \left| \frac{d^2y}{dx^2} \right|_{\text{max}} = \frac{c}{a^2}
   \]
   The maximum curvature of the ellipse, when the major axis is vertical, is:
   \[
   \kappa_{\text{ellipse}} = \frac{b}{a^2} = \frac{a}{a^2} = \frac{1}{a}
   \]
   For the ellipse to fit inside the sine curve at the tightest point, the curvatures must match:
   \[
   \frac{c}{a^2} = \frac{1}{a}
   \]
   Solving for \(c\):
   \[
   c = a
   \]

Therefore, the relationship between \(a\), \(b\), and \(c\) is:
\[
a = b = c
\]

The final answer is \( \boxed{ a = b = c } \)