Divide the space into two parts, above and below the plane where the head is located (the upper part of the head is divided into vertical direct rain area, and the lower part is divided into two directions)

(1) Rain is blowing from the head:

1. The whole raindrop under the head can be regarded as motionless, and people are bumping upwards, so the rainfall in the front and back direction has nothing to do with the speed (W 1=abd).

2. For the overhead part (vertical direct rain surface), the rainfall is inversely proportional to the speed. (Note: No matter which direction the rain falls, the amount of rain falls on people's heads is equal when they are still) (For the overhead part, the amount of rain per unit time and area is W0=ucos? Running distance d (m) is t=D/V, rainfall W2= W0tbc= ucos? Dbc/V)。

(2) the rain is blowing from behind:

1. In the front and back direction, if the speed of people and rain is the same (ideal state), it will not rain around.

2. The overhead part (vertical direct rain surface) is the same as "(1)".

(3) When the rain has a lateral (left and right) speed, just add the rainfall on the side after the discussion of (1)(2) (the solution is the same as above).

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