How do I Sum Vectors to Find Direction?

[GeoFamVS]

In a GeoCache we are trying to find, there were 4 clues with offset information.

Final clue is to sum the vectors?? How do you do that Math? Please help.

 

#1 110m 320 deg

#2 130m 200 deg

#3 60m 60 deg

#4 120m 130 deg

 

I'm supposed to sum thise, find a new co-ordinate, from the last location - which I did find...

Any ideas?

[+dfx]

Easiest way is to do it one by one, either on your GPS or through an online service such as http://www.gpsvisualizer.com/calculators

 

Take the origin waypoint, do the first projection, you get an end waypoint. Take that end waypoint and use it as origin waypoint for the second projection, and so on.

[GeoFamVS]

Thanks, I will take a look at that site - Wondering this, the description did say something about not being able to do one at a time, as these are around a lake, you would end up in the water ! does that make sense? does doing them one by one still work then? sorry, we are newbies to Geocaching... Thx

[+ecanderson]

Easiest way is to do it one by one, either on your GPS or through an online service such as http://www.gpsvisual...com/calculators

To the OP: Explicitly, at that link, using the "Find the coordinates at a given distance and bearing", repeated 4 times using the results of each prior run to start the next one, will do the trick.

Over short distances of the sort you need, it's not a tough calculation, but you'd still need to find out the approximate feet per 0.001 minute at your latitude. The online calculator is a lot easier!

Edited September 23, 2011 by ecanderson

[GeoFamVS]

Easiest way is to do it one by one, either on your GPS or through an online service such as http://www.gpsvisualizer.com/calculators

 

Take the origin waypoint, do the first projection, you get an end waypoint. Take that end waypoint and use it as origin waypoint for the second projection, and so on.

 

Thanks, I will take a look at that site - Wondering this, the description did say something about not being able to do one at a time, as these are around a lake, you would end up in the water ! does that make sense? does doing them one by one still work then? sorry, we are newbies to Geocaching... Thx

[+dfx]

Thanks, I will take a look at that site - Wondering this, the description did say something about not being able to do one at a time, as these are around a lake, you would end up in the water ! does that make sense? does doing them one by one still work then? sorry, we are newbies to Geocaching... Thx

 

Yeah it will still work. I'm guessing that what the description means is that doing the first projection, then actually going to that waypoint, then doing the second projection from there, etc, would not work. But you don't need to do that, you can do all the projections at home and then only go to the final waypoint.

[+ecanderson]

Thanks, I will take a look at that site - Wondering this, the description did say something about not being able to do one at a time, as these are around a lake, you would end up in the water ! does that make sense? does doing them one by one still work then? sorry, we are newbies to Geocaching... Thx

They are saying that you cannot use your GPS to do them one at a time because one or more of the endpoints is somewhere in the lake. So without a boat, you'd need to calculate all of them and find the end result, not try to walk it with your GPS!

[GeoFamVS]

sorry - removed the spoiler info...

Edited September 23, 2011 by GeoFamVS

[+dfx]

Alright, now please edit your posts and remove the actual numbers, because now it's just plain spoilers.

[GeoFamVS]

I can't seem to get the calculator to work...

 

If the location is N043degrees 44.726 and W079 degrees 44.041 - how do you enter that into the online calculator? http://www.gpsvisualizer.com/calculators

 

N43.44.726, W79.44.041? doesn't seem to be accurate when I map it? what am i doing wrong?

Edited September 23, 2011 by GeoFamVS

[+ecanderson]

I can't seem to get the calculator to work...

 

If the location is N043degrees 44.726 and W079 degrees 44.041 - how do you enter that into the online calculator? http://www.gpsvisual...com/calculators

 

N43.44.726, W79.44.041? doesn't seem to be accurate when I map it? what am i doing wrong?

You can enter it in this format "43 44.726,-79 44.041" but don't be surprised when it spits out the next value in a format like "43.51975672048, -79.19796920745".

Use that in the next iteration.

[+Red90]

At those distances, simple trig will suffice.

 

X = D cos A

Y = D sin A

Sum the X and Y from each projection for final location. Basic vector math.

 

For instance vector 1:

110 m at 320 degrees.

X = 110 cos 320 = -70.7 m (- means west)

Y = 110 sin 320 = 84.3 m (+ means north)

 

Do this for each one and add up the X and Y values. X (total) = X1 + X2 + X3 + X4...... and then you have the distance from the start to the end.

Edited September 23, 2011 by Red90

[+taeke]

If you have an Android phone, you can use the calculator of GCC (Coordinates - Projection), and work out all your coordinates till you have the end-coordinate.

[+ecanderson]

Just a generic reminder to all (not you in particular, Red) when talking distances. I did not check the user's location. But at Houston latitude vs. the Canadian border, there's a big difference in what 0.001 of E/W longitude means in actual feet, even over fairly short distances. Fortunately, the OP's example doesn't get involved in this, but for some even simpler calculations for puzzles, this makes a difference.

 

While feet / 0.001 minute is consistent when talking latitude (I use 6.076 feet per 0.001 N/S for this), the E/W distance represented by 0.001 increases toward the equator, and gets pretty squished as you head north. I round to 4.7 feet per 0.001 minute in my area for short distances.

 

Here's a few good rules of thumb about what a degree of longitude means at various latitudes:

 

@30º, 0.001 E/W on the GPS = 5.274 feet

@35º, 0.001 E/W on the GPS = 4.992 feet

@40º, 0.001 E/W on the GPS = 4.669 feet

@45º, 0.001 E/W on the GPS = 4.312 feet

@49º, 0.001 E/W on the GPS = 3.995 feet

 

So if a person were thinking on the order of 0.100 on the GPS, the difference E/W between a Houston calculation (527') and Toronto (400') would be a miss of 127'.

Edited September 23, 2011 by ecanderson

[39_Steps]

Fortunately, the OP's example doesn't get involved in this, but for some even simpler calculations for puzzles, this makes a difference.

 

[snip]

 

So if a person were thinking on the order of 0.100 on the GPS, the difference E/W between a Houston calculation (527') and Toronto (400') would be a miss of 127'.

If I were to try to use anything more exotic than what I remember of plane geometry to solve the OP's localized projection problem, no telling what the results might be.

 

I did pick a relatively uninhabited spot at about N 113 latitude to project those cache vectors from the same coordinates on both Delorme Topo9 and PN-60 map screens.

 

There was an 18 meter discrepancy between the results, as determined by closing the polygon back to the origin.

 

Even so, the results were probably better than trying Windows Calculator in "scientific" mode, as I have tried in the past. There's a litle rounding down going on with each of these utilities. Your coordinates may vary.

[+taeke]

Just a generic reminder to all (not you in particular, Red) when talking distances.

 

I'm not a scientist, so I might be wrong, but I don't think that matters when you do a calculation on different latitudes, as the angles are different of the lines you use a reference. Correct me if I'm wrong.

[jholly]

 

I did pick a relatively uninhabited spot at about N 113 latitude to project those cache vectors from the same coordinates on both Delorme Topo9 and PN-60 map screens.

 

I'm curious, where did you find a place that was N 113 latitude?

[GeoFamVS]

Thanks for everyone's help.. unfortunately I'm still lost Tried figuring out how this:

X = D cos A

Y = D sin A

Sum the X and Y from each projection for final location. Basic vector math.

 

For instance vector 1:

110 m at 320 degrees.

X = 110 cos 320 = -70.7 m (- means west)

Y = 110 sin 320 = 84.3 m (+ means north)

 

Do this for each one and add up the X and Y values. X (total) = X1 + X2 + X3 + X4...... and then you have the distance from the start to the end.

 

But not sure how to enter this info into a calculator that has SIN etc.

ALso - once you have the total sum - how do you put that into a meaningful direction and distance - I am using the geocaching.com app on iPhone and have a road garmin GPS if needed...

Sorry - we're still newbies.. but enjoy this hobby - this one is very challenging! and I have a travel bug to place in it once i find it !!!

[+EScout]

Some GPSrs will let you project from a saved waypoint. Just do this for each of these distance/bearings sequentialy. You have to know if you should be using true or magnetic north reference.

There are also some apps that can do this on computer or phone.

[John E Cache]

In a GeoCache we are trying to find, there were 4 clues with offset information.

Just curious. Were the "offsets" North/South East/West offsets. If so, couldn't you just add the offsets.

[GeoFamVS]

Eg N43 44.504

W 79 44.220

Offset 110 m. 320 deg m

[39_Steps]

 

I did pick a relatively uninhabited spot at about N 113 latitude to project those cache vectors from the same coordinates on both Delorme Topo9 and PN-60 map screens.

 

I'm curious, where did you find a place that was N 113 latitude?

I must have been out on that spacecraft that's due to reenter, crash and burn "real soon now".

 

It was in some base in southern Arizona named after some person by the name of Barry.

[+ecanderson]

Just a generic reminder to all (not you in particular, Red) when talking distances.

 

I'm not a scientist, so I might be wrong, but I don't think that matters when you do a calculation on different latitudes, as the angles are different of the lines you use a reference. Correct me if I'm wrong.

The reminder was NOT specific to the OP's situation, as I had hoped to make clear. Shooting a bearing at a known distance with a GPS will create no issues, and there are no issues of the sort I mentioned using the solutions we presented to the OP.

 

But if a person sits down and attempts to compute locations in feet (X feet one way, Y feet the other) using simple angles and distances, you can run into large errors quickly by making bad assumptions about how much distance 0.001 minute respresents E/W. Again, it was a somewhat generic reminder about calculating solutions to certain puzzles. For example, if you find yourself solving a puzzle that requires you to determine so many feet X and Y from a given point, it's nice to be able to quickly reduce that to a set of coordinates from the starting point and just navigate to the coordinates. You need to know how many feet per 0.001 (the common units in our case) to do that. There are other types of puzzles and multis that also require that knowledge. It depends upon how the vectors are presented.

Edited September 23, 2011 by ecanderson

[+ecanderson]

Eg N43 44.504

W 79 44.220

Offset 110 m. 320 deg m

Just use that nifty online calculator that was presented to you earlier (http://www.gpsvisualizer.com/calculators)

 

At the bottom, in the section called "Find the coordinates at a given distance and bearing", enter as follows:

 

Starting Lat/Long: 43 44.504,-79 44.220

Distance: 110 m

Bearing: 320

 

That takes you to 43.74249063442, -79.7378, which for useful purposes, is 43.74249 / -79.73780

 

If you need to project another from there, plug in the 43.74249,-79.73780 (five digits after the decimal are enough) with the distance and bearing given for the next point, etc etc. until you've done all 4.

 

If the format 43.74249063442, -79.7378 bothers you, understand it's quick to get from decimal degrees to the format we typically use:

 

43.74249 = 43 degrees and 60 x 0.74249 minutes, or

43 degrees, 44.549 minutes = 44º 44.549 in geocaching units.

[John E Cache]

You could try the method I posted for surveyor plot vectors.

 

http://forums.Ground...dpost&p=4793223

 

Hopefully, you can walk it without getting wet.