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Crosswind components
Can someone offer a non-mathematical EXPLANATION (as opposed to
DESCRIPTION) of why the speed of headwind and crosswind components of a wind add up to more than the speed of the wind? Thanks. |
#2
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"James L. Freeman" wrote in message om... Can someone offer a non-mathematical EXPLANATION (as opposed to DESCRIPTION) of why the speed of headwind and crosswind components of a wind add up to more than the speed of the wind? The same reason the opposite and adjacent sides of a right angle triangle add up to more than the hypotenuse. I guess that's mathematical, but it is an explanation... Regards Andrew |
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"James L. Freeman" wrote in message
om... Can someone offer a non-mathematical EXPLANATION (as opposed to DESCRIPTION) of why the speed of headwind and crosswind components of a wind add up to more than the speed of the wind? For the same reason that it's a shortcut if you can walk diagonally from one corner of a city block to the opposite corner, rather than following the roads around the block. --Gary Thanks. |
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You shoulda stayed awake during trigonometry...
MikeM James L. Freeman wrote: Can someone offer a non-mathematical EXPLANATION (as opposed to DESCRIPTION) of why the speed of headwind and crosswind components of a wind add up to more than the speed of the wind? Thanks. |
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It's because they are separate definitions of the crosswind and are not
additive. -- B-58 Hustler History: http://members.cox.net/dschmidt1/ - "James L. Freeman" wrote in message om... Can someone offer a non-mathematical EXPLANATION (as opposed to DESCRIPTION) of why the speed of headwind and crosswind components of a wind add up to more than the speed of the wind? Thanks. |
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Because wind has speed and direction. You cannot just add the numbers
to get the total. You have to do a vector sum (considering direction and speed). (James L. Freeman) wrote in message . com... Can someone offer a non-mathematical EXPLANATION (as opposed to DESCRIPTION) of why the speed of headwind and crosswind components of a wind add up to more than the speed of the wind? Thanks. |
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In article ,
Casey Wilson wrote: (James L. Freeman) wrote in message .com... Can someone offer a non-mathematical EXPLANATION (as opposed to DESCRIPTION) of why the speed of headwind and crosswind components of a wind add up to more than the speed of the wind? I'll give it a go (use to tutor math in college but that was a while ago ^_^). This works best with pictures but here's teh basic outline. Imgine you have a block on the floor. Two guys (A and B) are going to push on the block. Since we're doing this wihtout pictures let's imgine you're looking at the block from teh top down and it can go North/south/east/west over teh floor. 1. A pushes on the south side of the block with 5 units (of whatever), B pushes on the south side of the block with 5 units of force. The result the block moves to the north with 10 units of force. 2. A pushes on teh south side of the block with 5 units, B pushes on the north side of the block with 5 units. Result, teh block doesn't move at all. But Both A & B are pushing with 5 units. 3. A pushes on the south side of teh block, B pushes on the east side f teh block. Result is the block moves north-east. That sounds ok right? But wait a mintue, in example 3 if only A was pushing then the block would move north and only north. And if only B was pushing the block would move east and only east. The block moves north-east because both are pushing. In example 2 the block didn't move at all, because A and B are pushing in opposite directions with equal force. But What if A pushing on the south side of rh block with 10 units and B pushes on teh north side with 8 units? Well the block would move north, at 2 units ofr force. From the blocks POV having A push from teh south at 10 & B push from teh north at 8, is THE SAME THING as just having someone push at 2 units of force. This is know as vector addition. You need to take teh scaler (number of units of push) and the vector (direction) into account to do vector addition. So if I told you that A was pushing at 10 units and B was pushing at 10 units, and asked what direction is the block moving what would you say? Well you 'should' say "I don't kow." Because you don't know what direction A and B are pushing. The above examples are easy because A nd B are pushing in opposite direttions. It's a little harder to think of it when A and B push between 0-180 degrees. The math is easy (sin/cos stuff or use a cross wind diagram). It's the same thing thou. Let's say you want to land on runway 36 (to make life easy). The wind is 045 @ 20. Well just like teh block having the win at 045 @ 20, is the same as having two winds that are pushing the plane at the same time. One wind is coming from 360 and the other from 090, the end result is a single wind from 045 (yes I know wind doesn't work like that work with me here). The reason we 'spilt the wind up' is because the 'peice' of wind coming from 360 doesn't concern us, but that peice from 090 does, as it'll blow us off course when landing. So it's the reverse of the block examples, we have one force at some odd angle, and we spilt it into two force: head wind and cross wind. Because these forces have strength (number of units) and direction (in this case 045) we can't just add/subtract like number. Just like the block example above where the block was moving forward at 2 units, wind acts the same. Landing on 360 with a wind from 360 @ 10 is all head wind, 015 @ 10 a little cross wind, 080 @ 10 almost all cross wind. So the cross wind component varies with direction, and you can't just add/subtract the units to get there. Now find any url about trig and look up sin and/or cos and this will make a lot more sense (I hope). But you asked why and not how to calculate it This stuff is actually pretty easy to do if you have a scienctific calculator. Sorry the above is so long ... |
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"Casey Wilson" wrote in message ...
(James L. Freeman) wrote in message . com... Can someone offer a non-mathematical EXPLANATION (as opposed to DESCRIPTION) of why the speed of headwind and crosswind components of a wind add up to more than the speed of the wind? Thanks. "Andrew Sarangan" wrote in message om... Because wind has speed and direction. You cannot just add the numbers to get the total. You have to do a vector sum (considering direction and speed). I don't think that's the answer. When I run a wind problem on my Whiz-Wheel, the resultant IS a vector solution and the crosswind is only a fraction of the wind velocity. Something makes me want to say it is the cosine of the angle off the wing. Where you do think the cosine comes from? It is a consequence of adding two vectors. A 10 knot wind from the northwest has a westerly component of 10cos(45) = 7knots and a northerly component of 10cos(45) = 7knots. Alternatively, if you have 7 knots from the west and 7 knots from the north, you wind up with a total of 10 knots from the northwest, not 14 knots. |
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