VANCOUVER ISLAND WINDTALK

correct me if i'm wrong, but i calculate that for every 1.58 km you travel "up" island, west is 1 km, and the rest is north

Depends what you mean by 'rest'. It isn't 0.58. N and W are two sides of a right angle triangle with 'up island' being the hypoteneuse. So remember the old A squared + B squared = C squared? Playing with Google Earth I measure about 450 km from Victoria to about Cape Scott. Keeping the Cape Scott Lat and moving to the long of Victoria I get 276 km N; and doing the same for W, I get 363 km.

Squaring the N and W and adding them together then taking the square root I get 456 km, pretty close to the measured straight line. So that seems to work.

So back to your numbers, for every km W you go 276/363 km N (about 0.76 km N); and 0.76 squared plus 1 squared is 1.5776 and taking the square root that is 1.25 km 'up island'. Depends which parts of the island you measure - going up the west coast would be different numbers (like to Tofino).

I expect my math will be corrected by some of the profs and programming geeks on here!

holy crap I sure hope we get more wind soon

ME TOO !

No wind likely for more than a week guys, better to brush up on math and geography!

It depends on the angle of the island. Here is a chart that gives the distance traveled north (sin of the angle) and west (cosine of the angle) for every 1 km traveled "up island" as a function of the angle (with respect to the horizontal.) I don't have a protractor to measure the angle but it looks like 45 or 50 degrees to me. MF4's analysis is for 45 degrees. SORRY I HAVE BEEN CORRECTED. I DIDN'T READ CAREFULLY. THE ANGLE HE USED IS 37.24 degrees I've added a line for MF4's analysis. I still don't have a protractor so I can't measure the angle but I guess it is somewhere between 37.24 and 50 degrees.

Angle......N .......W
40......0.64......0.77
45......0.71 ......0.71
50......0.77 ......0.64
37.24.......0.61.......0.80

Going back to the 1.58 km travel, you multiply 1.58 by the number in the table above to get the result. This is the table.

Angle.......N .........W
40.......1.02......1.21
45.......1.12......1.12
50.......1.21......1.02

The angle looks like it is roughly 50 degrees, so the statement "i calculate that for every 1.58 km you travel "up" island, west is 1 km, and the rest is north" is correct except for "the rest" part.

We need wind.

Geez, Mike, mine isn't for 45 degrees!!; I didn't mess with a protractor, I used Google Earth which is actually curved geometry and I assumed flat; but the island is 363 km 'west' and only 276 km 'north', so that can't be 45 degrees which would be the same both dimensions.

The interesting thing is that the island is much more east-west than it is north-south which I think was Wavos' original point.

BTW Mike, angles used on maps or in surveying are usually measured from north, not 'horizontal' (using 0 as west)

We surely do need wind!

Thanks for correcting me MF4. See my previous post for an apology.

Have a look at http://www.gmap-pedometer.com/?r=1461070.

It's a little rough but I make it 398 km from Victoria to Port Hardy. 246 km north/south and 305 km west on the right angle triangle.

That would make the angle 38.9 degrees.

Doohhh! get back to work ....

holy crap I sure hope we get more wind soon

OK MTM, you do qualify it as 'a little rough' and you pick Port H which is maybe a little over to the 'north' side of the island, but I want to point out your basic measurements are a bit off. I went to your web site there and your triangle base isn't east-west - I pulled out my Douglas protractor and checked your base compared to the 49th parallel on the same map, and they should be parallel but aren't. You are probably using the edge of the screen as 'north-south' but of course map projection starts to play a role here. Depending on the projection used to make the 2-D map, you can get the angles correct but the distance will be a bit off, or the distance right but the angles will be off, or both off but only by a bit, etc. etc. Thats why the distances in Google Earth using the lats and longs to stay parallel are a better way. I think the measurements in Google Earth are based on the Great Circle.

Don't worry Kus, I think we can keep this thread going for some time yet....!!!!!

Holy crap is right KUS! MF4 is correct though. Nothing in my life is at right angles!

I told you MF4 I didn't have a Mike Protractor, let alone a "Douglas Protractor"!

Good thing I can see the far side at Nitinat cause I would end up in Hawaii.

Now lets have a look at the elevation angle change. Here I am approximating the elevation angle, not accounting for the change in air density, to be ....

http://whom.co.uk/html/protract.htm

Probably the cheapest thing on sale at the "VIP Pilots Centre"
http://www.boutiquescanada.com/Magasin/ ... 758.2.html

Nobody should leave home without one! As for setting sail across the trackless wastes of Nitinat Lake without knowing basic navigation and pilotage - reckless in the extreme!

I got the software, but you guys got the know how...

I used oziexplorer navigation charts to set the waypoints on this picture, 1,2 and 3



Waypoint 3 is due west of waypoint 2, and waypoint 1 is due north of waypoint 3




These are the distances between the waypoints

NOT to complicate things even more for you guys, but the direction and distance measured on a 2D map are very difficult to do accurately. It depends on the map projection you have.

For instance, you can only have accurate distance on an equidistant map projection FROM THE REFERENCE POINT of the map, i.e. distances from only one spot are accurate, everything else is garbage.

To add to the confusion, to have accurate angles measured from a map, you need an azimuthal map projection, which has it's own reference point restriction.

Unfortunately, to the best of my knowledge, you cant have a perfectly azimuthal and equidistant map projection (unless you take measurements from a sphere, have fun with that geometry), so this exercise will only ever give you an approximation!

p.s. I too hope it gets windy soon. haha