rain and umbrella

[sidharath]

The rain is falling down vertically and the person holding umbrella is moving with very quick pace. To avoid getting wet should the umbrella held in vertical or oblique direction.

[Ophiolite]

Oblique.

[CaptainPanic]

You should take the person as the stationary reference point and calculate the angle of the rain in relation to that reference point. Then hold the umbrella at that angle right over the part of the body that you wish to keep dry (in many cases you cannot keep the whole body dry).

[studiot]

The rain is falling vertically so there is no wind.

 

In these circumstances you should hold the umbrella vertically.

 

The umbrella cast a 'dry shadow' over the holder which is maximal at the vertical umbrella position, whether the umbrella is moving or stationary.

 

At any other angle the shadow is oblique and equal to the projection of the umbrella area onto the horizontal.

 

This is very much like the question,

 

"Is the normal reaction at an angle when on object moves over its support?"

Edited September 30, 2013 by studiot

[CaptainPanic]

The umbrella cast a 'dry shadow' over the holder which is maximal at the vertical umbrella position, whether the umbrella is moving or stationary.

 

A person who does not walk has only an top surface that will get wet (essentially the head and shoulders).

A person who is walking, will have a top and a frontal surface (now also including legs, chest and arms).

 

The horizontal umbrella will cast a maximum 'shadow' over the horizontal surface area, but will cast a rather poor 'shadow' over the vertical areas. As the speed increases, the amount of rain that a person will catch with the vertical surface areas will increase. And therefore you must attempt to increase the overall surface area, which you will achieve by holding the umbrella at an angle. Which angle was explained in my previous post.

[studiot]

Sorry I can't agree with this.

 

No rain can fall directly down through the horizontal projection of the umbrella, wherever it is.

The umbrella creates a 'dry zone' immediately in front of the walker.

The walker is always walking into this zone.

Since the horizontal position maximizes the shadow it maximizes the size of this dry zone.

The umbrella travels with the walker.

 

So yes, the walker is walking into the rain, but the only rain that she can encounter has already fallen below the level of the umbrella, before the umbrella has arrived.

 

So it becomes a race between the speed of the walker and the speed of falling drops.

 

Yes the vertical projection of the umbrella will push the drops in the upper level out of the way.

But this projection must be small to keep off the drops still falling vertically above the walker.

The walker will still catch any drops that have already fallen below this level on the lower part of their body, shoes etc, if he walks fast enough.

The risk of that is minimised by maximizing the dry zone and therefore the horizontal projection.

 

This is unlike the situation of an open topped sports car, which is going much faster, so a (near) vertical windscreen will push the drops in the upper half away and the car's speed will carry you past the vertically falling drops before more can fall. Further the driver is protected at low level by the bodywork.

Edited September 30, 2013 by studiot

[michel123456]

Sorry I can't agree with this.

 

No rain can fall directly down through the horizontal projection of the umbrella, wherever it is.

The umbrella creates a 'dry zone' immediately in front of the walker.

The walker is always walking into this zone.

Since the horizontal position maximizes the shadow it maximizes the size of this dry zone.

The umbrella travels with the walker.

 

So yes, the walker is walking into the rain, but the only rain that she can encounter has already fallen below the level of the umbrella, before the umbrella has arrived.

 

So it becomes a race between the speed of the walker and the speed of falling drops.

 

Yes the vertical projection of the umbrella will push the drops in the upper level out of the way.

But this projection must be small to keep off the drops still falling vertically above the walker.

The walker will still catch any drops that have already fallen below this level on the lower part of their body, shoes etc, if he walks fast enough.

The risk of that is minimised by maximizing the dry zone and therefore the horisontal projection.

 

This is unlike the situation of an open topped sports car, which is going much faster, so a (near) vertical windscreen will push the drops in the upper half away and the car's speed will carry you past the vertically falling drops before more can fall. Further the driver is protected at low level by the bodywork.

You wrote: "So it becomes a race between the speed of the walker and the speed of falling drops."

 

An umbrella does not absorb the water drops but displace them at the perimeter, creating a cylinder of water around the walker.

 

So we are not talking about the falling drops from the sky only, but mainly the falling drops from the umbrella's perimeter.

 

Which are quite slow: they are in simple free fall with negligeable initial speed at umbrella's edge.

 

Also, the drops at umbrella's edge have now the same horizontal speed with the walker.

 

In this case of figure, the horizontal umbrella gives the largest cylinder of water and Studiot is correct.

 

In practice the drops from the edge do not fall with the same horizontal speed with the walker because the walker creates "wind" by its own displacement and the drops are falling inclined as observed by the walker. This fact does not change whether the umbrella is inclined or not (it changes slightly the height of the edge but who matters). Your head is protected by the umbrella but not your pants because your feet will get very close the water cone if you walk quickly.

 

And also in practice it is almost impossible for the walker to keep the umbrella vertical because of the "wind" he creates being considered the shape of the umbrella.

 

IOW in practice, the walker is obliged to keep the umbrella inclined not regarding at all the issue about the drops. The important thing is to keep your head dry.

Edited September 30, 2013 by michel123456

[sidharath]

Thank you very much indeed ophiolite captainpanic studidiot for your reply.i shall be highly thankful to you if you let me know whether the rain drops strike the surface of the umbrella in vertical or oblique direction when the walker is moving very briskly. I am a little confused

[studiot]

Here is my analysis of the situation. The notation provides a basis for futher discussion.

 

Here is some explanation.

 

OP is the projected length of the umbrella on the horizontal. This blocks all rain from passing.

As the front edge, B, moves forwards it covers rain that has already fallen below the level of the umbrella.

This rain continues to fall vertically, impacting the ground between C and D.

C is the point where rain falling when the umbrella edge was at A strikes the ground.

These points create a diagonal line BC, separating the area under the umbrella into two zones.

Quad OBCP where is it totally dry and triangle, BCD where it is still raining.

 

Thus any walker behind line AC cannot get wet under any circumstances

 

Further, although there is rain in triangle CED, none of this is accessible to the walker, since it will have fallen to the ground by the time the walker arrives.

 

Thus only the rain in triangle BEC could ever hit the walker.

 

Now since BD is constant the longer AB is the smaller the slope of CB and therefore the smaller the triangle BCE.

 

But the condition for making AB as long as possible is for it to be horizontal.

 

This answers the original question, but an interesting further exercise is to superimpose the area within BCE that the walker is going fast enough to enter. This is the point of line A'B'E'C' and the velocities.

 

 

Thank you very much indeed ophiolite captainpanic studidiot for your reply.i shall be highly thankful to you if you let me know whether the rain drops strike the surface of the umbrella in vertical or oblique direction when the walker is moving very briskly. I am a little confused

 

When the rain is vertical and the umbrella horizontal the they meet at 90 degrees.

If the umbrella is now tilted at x from the horizontal, the rain and umbrella are now at (90-x) to each other.

Edited September 30, 2013 by studiot

[michel123456]

Here is my analysis of the situation. The notation provides a basis for futher discussion.

umbrella1.jpg

 

Here is some explanation.

 

OP is the projected length of the umbrella on the horizontal. This blocks all rain from passing.

As the front edge, B, moves forwards it covers rain that has already fallen below the level of the umbrella.

This rain continues to fall vertically, impacting the ground between C and D.

C is the point where rain falling when the umbrella edge was at A strikes the ground.

These points create a diagonal line BC, separating the area under the umbrella into two zones.

Quad OBCP where is it totally dry and triangle, BCD where it is still raining.

 

Thus any walker behind line AC cannot get wet under any circumstances

 

Further, although there is rain in triangle CED, none of this is accessible to the walker, since it will have fallen to the ground by the time the walker arrives.

 

Thus only the rain in triangle BEC could ever hit the walker.

 

Now since BD is constant the longer AB is the smaller the slope of CB and therefore the smaller the triangle BCE.

 

But the condition for making AB as long as possible is for it to be horizontal.

 

This answers the original question, but an interesting further exercise is to superimpose the area within BCE that the walker is going fast enough to enter. This is the point of line A'B'E'C' and the velocities.

Some observations:

1. you should draw your diagram at scale, with real umbrella diameter and real height, real velocities.

2. draw a human walking inside.

3. draw the rain behind

4. in your diagram, the drops are only those falling directly from the sky, at maximum velocity (how much is that?). You should also put the other drops falling from the perimeter, at free fall beginning with v=0 at the perimeter, that is much slower than the other ones. If I am not abused, those slow drops will produce a CB slope much less that will intersect with the walker's feet.

[studiot]

1. What purpose does this serve? My objective is calculation not drafting perfection. How do I draw a velocity to scale? This is an instantaneous picture. The velocities are linked by formulae.

I have modelled my human as a vertical straight line, since this line will meet the rain before any other part of the human.

I have no idea what you mean by draw the rain behind.

Again I have no idea what you mean. Raindrops falling from the perimeter ? All the raindrops have reached terminal velocity and are falling at the same constant speed. These were the explicit conditions stated by the OP. Water splashes, water from another source, etc are all outside the scope of this model. But then so is the slope of the ground, the partial evaporation of rain drops and other phenomena present in the real world.

Edited September 30, 2013 by studiot

[michel123456]

1. What purpose a drawing at scale? If you don't draw an exact diagram you may get the wrong conclusions, look here for example. Velocity at scale: imagine the rain falls at excessively great speed (at SOL), then the rain always fall vertically because the walker's velocity is negligeable compared to the velocity of rain. On the other hand, if rain's velocity is near zero, then the rain falls nearly horizontally as observed by the walker and he gets wet all over. IOW the relation of vertical drops velocity and walker's velocity gives the angle in your diagram.

2. say yes, a vertical straight line that has a specific height.

3. the rain behind falls in 2 ways A the rain coming directly from the sky, falling at great velocity (nearly vertically for any walker's speed) and B the rain slipping from the umbrella with zero initial velocity at umbrella's edge falling at an angle observable on the diagram.

4. yes, drops falling from the perimeter. The object of the exercise is finally to find the largest dry volume, not the largest projection on the ground.

[sidharath]

thank you very much studidiot for the trouble you have taken in sending detailed reply. You have mentioned correctly that raindrops strike the umbrella vertically whether it is moving or stationary. I was confused because in some sub standard text it was mentioned that when the umbrella is moving the velocity of the falling drops is the resultant of velocities of drops and that of walker acting at 90 angle hence the rain drops strike the umbrella in oblique direction as a consequence of which umbrella should be held in oblique direction for maximum protection

[michel123456]

These are my thoughts:

 

 

 

[studiot]

Good job you finished your post, michel, before I finished mine. I can give a better reply.

 

The original post asked for the optimum placement of the umbrella. Your sketches do not show this.

 

For example the walker in the third sketch would be better protected at the rear if he held the umbrella higher and more nearly vertical.

 

As I said at the end of post#9, my analysis answered the specific question posed. However it can also be used and adpated to answer all sorts of variations and 'what if' questions depending upon the variation of velocities, heights, size of umbrella, and so on.

 

It can show that the faster the walker goes the further line AC penetrates under the canopy so provides a limit where AC would actually coincide with OP.

It can show that the closer the umbrella is to the walker's head the better he is protected, ie he should not hold it high in the air.

 

You diagram shows that the curvature of a real umbrella provides a measure of slant protection without tilting the umbrella in any case.

Edited October 1, 2013 by studiot