Illustrative Example

As an illustrative example, the physical and hodograph plots of a common cliché in aerodynamics, potential flow about a cylinder, are given. Figure 1 presents the pathlines for such a flow in the physical plane, and Figure 2 presents the pathlines transformed to the hodograph plane. (Of course, since we are assuming steady flow, the pathlines coincide with the streamlines.)

  

Figure 1: Streamline plot of potential flow around a cylinder in the physical plane.

  

Figure 2: Streamline plot of potential flow around a cylinder in the hodograph plane.

Both figures show the same pathlines. Whereas in the physical plane, a pathline traces the physical location of a particle as it travels along; in the hodograph plane, a pathline traces the velocity of a particle. The axes of this hodograph plane are the x and ycomponents of velocity (u and v).

To see how this works, imagine first that you are a particle on the pathline that ends at the stagnation point of the cylinder. Far from the cylinder, your velocity is essentially equal to the free-stream velocity $V_\infty$. There is no vertical component to your velocity. In the hodograph plane, this is represented by a point on the u-axis (since there's no v component) very slightly less than $V_\infty$.

As you approach the cylinder, you begin to slow down. Your velocity is changing, therefore your position in the hodograph plane is changing also: you are approaching zero velocity along the u-axis. At the stagnation point, where your path ends, your velocity is zero. In the hodograph plane, you are at the origin. Thus the stagnation pathline maps to a straight line from $(V_\infty,0)$ to (0,0) in the hodograph plane.

Now, imagine that you're a particle very slightly above the stagnation streamline. Far from the cylinder, your velocity is essentially $V_\infty$, but you do have a very small vertical component. In the hodograph plane, you are at a point very close to $(V_\infty,0)$, but you're not exactly on the v-axis.

As you approach the cylinder, you begin to decelerate. Because you are close to the stagnation pathline, you get very close to the stagnation point. You are traveling very slow at this point, but have finite velocity. The vertical component of your velocity has increased slightly. On the hodograph plane, you are at a point near the origin, slightly above it and to the left.

Due to the high pressure at the stagnation point, you will then be accelerated upward, on your way around the cylinder. Initially, your acceleration is mostly vertical, so v component begins to rise. As you travel around the cylinder, your direction changes from mostly vertical to horizontal above the apex. Above the apex, the vertical velocity is zero. Somewhere between the apex and the stagnation point, your vertical velocity reaches a maximum.

In the hodograph plane, your pathline from near stagnation to above the apex looks like the top half of an oval. It starts at the near-stagnation point slightly above and to the left of the origin, and arcs over to a point on the u-axis slightly less than $2V_\infty$, which corresponds to the point above the apex.

After the apex, you go through mostly the reverse of what you had gone through up to this point, except that your vertical velocity is now negative. Therefore, in the hodograph plane, your pathline from the apex is the bottom of an oval, starting from the apex point, and arcing under to a point just below and to the right of the origin, which is near the aft stagnation point. Then, your pathline returns to the point $(V_\infty,0)$ as you accelerate back to free stream velocity.

Now, imagine that you're a particle on a pathline quite a bit above the stagnation pathline. You will experience acceleration similar to the previous particle, but to a lesser extent. You will decelerate, but not to almost zero. You will be going fastest right above the cylinder. However, your deviation from the free stream condition will be less. Your hodograph pathline would be a smaller oval-shape inside the larger. In general, we can say that the farther a pathline is from the stagnation pathline in the physical plane, the smaller your oval is in the hodograph plane.

Finally, imagine that you are a particle slightly below the stagnation pathline. You are now on a course to travel under the cylinder. In the hodograph plane, your pathline coincides with the particle whose pathline was slightly above the stagnation pathline, exept that it is in the opposite direction. Note that two diametrically opposite points (about the origin) in the physical plane have identical velocities. These two will map to the same point in the hodograph plane. Thus, the hodograph transformation is not one-to-one.