Well, the vertical component of wettining is a function of the terminal velocity of the rain (a function of drop size), the density of the rain, and the vertical cross-section of the human subject. The horizontal component of wettining is a function of the relative horizontal velocity of the rain (delta of rain velocity and human velocity), the terminal velocity of the rain, the density of the rain, and the normalized horizontal cross-section of the human subject. At least I think that's how it works.

Each component is summed over the time interval to get total wetness, where the time interval is the distance to shelter divided by the speed of the human toward the shelter.

Assuming a 1:5 vertical:horizontal cross-section ratio and a relative horizontal velocity of about 15 mph for a typical (non-Olympic conditioning :? ) sprinting human in a 5 mph opposing wind, the horizontal component of wettining would be of around five times the vertical component of wettining. The speed of destination approach would be around 10 mph. This assumes a terminal velocity of 15 mph for the rain.

For a relative horizontal velocity of about 7.5 mph for a typical walking human in a 5 mph opposing wind, the horizontal component of wettining would be on the order of two and one-half times the magnitude of the vertical component. The speed of destination approach would be around 2.5 mph. Thus it takes four times as long walking as running to the destination in the rain. Under these assumptions running wetness is ~6w and walking wetness is ~14w, i.e. a bit over twice the wetness.

At least I think this is how it figures out under a reasonable set of assumptions. Note that the result is independent of the rain density.