I have a graph which has two lines.
The graph is generated from "random" data. I.e. not based on a formula or pattern. But there is always a point where the two lines intersect.
I'm trying to provide exact point (on x and y axis) where the lines cross.
Ive tried using slope/intercept formulas
And what if analysis.
However these methods only seem to work if the data is based on a formula or pattern.
I can sort the data and find the point where they are at their closest then take an average using data around that point to get an approximate match.
However is there any way to do this more accurately, or does the nature of my data(random data points) make this not possible using formulas/equations
Suppose that we have a scattered series of data X,Y randomly spaced (in the pic they are ordered, but this doesn't matter) and a line which shows the maximum limit we are considering for a sub-application.
Is there a combination of functions to choose the closest points below the orange line? I've tried with a MAXIFS + LOOKUP, but didn't solve anything.
The formula of your line is: y=1.17*x, so you create a helper column, containing a formula like:
=IF(1.17*A3-B3>0;1.17*A3-B3;100000)
This means: calculate the difference between the line and the point if that difference is positive. In case it's negative (which means that the point is above the line), then show a value which is that large that it won't be taken into account while calculating the minimum.
You drag this formula all over the column.
You calculate the minimum of that column (one of the easy ways to do this, is using the autofilter).
I have a large set of XYZ Cartesian points in Excel (some 40k actually) and was looking for a formula or macro to compare every point to every other point to get the distances between them.
The math to get the distance value between two 3D points is:
Distance=SQRT((X2 – X1)^2 + (Y2 – Y1)^2 + (Z2 – Z1)^2)
X1=the X value of the 1st point
X2=the X value of the 2nd point
Y1=the Y value of the 1st point
Y2=the Y value of the 2nd point
etc
Here is an example starting with 10 points:
http://i.imgur.com/U3lchMk.jpg
Would anyone know of a way to build this into Excel so that I can just copy the formula across the page to the horizontal limit? Or would you recommend a better way than using Excel?
As a secondary goal, I want to group the points into clusters that can connect by a distance lower than 2. But if I can accomplish the first goal, I can worry about the second later.
Actually, I was able to come up with the solution with a bit more research: i.imgur.com/9JL5Qni.jpg =SQRT(((INDIRECT("A"&$D2))-(INDIRECT("A"&E$1)))^2+((INDIRECT("B"&$D2))-(INDIRECT("B"&E$1)))^2+((INDIRECT("C"&$D2))-(INDIRECT("C"&E$1)))^2)
I have a data set that has height values every so often, like topography data in a straight line with GPS coordinates. I used the GPS coordinates and trigonometry to make a cumulative distance column. However, the distance between points varies. Sometimes its 10 cm sometimes its 13, sometimes its 40.
I would like to take the average height every 0.5 meters, but sometimes the distance column doesnt even land on a multiple of 0.5! This would mean my output column would be significantly shorter than my raw data column.
I think my main problem is I do not know what this process is called in order to Google it. Another problem is that the distances are irregular as mentioned above. Things I think may have something to do with it:
averageif?
binning? I do not want a histrogram though, just the data.
Thanks for the help and if you do not know the answer but at least know what I should be writing in the search bars that would be helpful as well. Thanks!
Perhaps this will work for you. I made up a series of distance vs height measurements and determined that a third order polynomial curve fit pretty well. (A different curve might best fit your real data, so you would have to alter the formula accordingly). I then used that formula to derive a set of new heights for the desired ditances at, in my example five unit differences.
The formula under Extrapolated heights is an ARRAY formula entered into all the cells at once. You select D2:D12, enter the formula in D2 and, hold down CTRL-SHIFT while hitting ENTER. If you did this correctly, you will have the same formula in each cell surrounded by curly braces {...}
Then you can decide how you want to "Average" the heights.
I'm currently drawing up a mock database schema with two tables: Booking and Waypoint.
Booking stores the taxi booking information.
Waypoint stores the pickup and drop off points during the journey, along with the lat lon position. Each sequence is a stop in the journey.
How would I calculate the distance between the different stops in each journey (using the lat/lon data) in Excel?
Is there a way to programmatically define this in Excel, i.e. so that a formula can be placed into the mileage column (Booking table), lookup the matching sequence (via bookingId) for that journey in the Waypoint table and return a result?
Example 1:
A journey with 2 stops:
1 1 1 MK4 4FL, 2, Levens Hall Drive, Westcroft, Milton Keynes 52.002529 -0.797623
2 1 2 MK2 2RD, 55, Westfield Road, Bletchley, Milton Keynes 51.992571 -0.72753
4.1 miles according to Google, entry made in mileage column in Booking table where id = 1
Example 2:
A journey with 3 stops:
6 3 1 MK7 7DT, 2, Spearmint Close, Walnut Tree, Milton Keynes 52.017486 -0.690113
7 3 2 MK18 1JL, H S B C, Market Hill, Buckingham 52.000674 -0.987062
8 3 1 MK17 0FE, 1, Maids Close, Mursley, Milton Keynes 52.040622 -0.759417
27.7 miles according to Google, entry made in mileage column in Booking table where id = 3
If you want to find the distance between two points just use this formula and you will get the result in Km, just convert to miles if needed.
Point A: LAT1, LONG1
Point B: LAT2, LONG2
ACOS(COS(RADIANS(90-Lat1)) *COS(RADIANS(90-Lat2)) +SIN(RADIANS(90-Lat1)) *SIN(RADIANS(90-lat2)) *COS(RADIANS(long1-long2)))*6371
Regards
Until quite recently, accurate maps were constructed by triangulation, which in essence is the application of Pythagoras’s Theorem. For the distance between any pair of co-ordinates take the square root of the sum of the square of the difference in x co-ordinates and the square of the difference in y co-ordinates. The x and y co-ordinates must however be in the same units (eg miles) which involves factoring the latitude and longitude values. This can be complicated because the factor for longitude depends upon latitude (walking all round the North Pole is less far than walking around the Equator) but in your case a factor for 52o North should serve. From this the results (which might be checked here) are around 20% different from the examples you give (in the second case, with pairing IDs 6 and 7 and adding that result to the result from pairing IDs 7 and 8).
Since you say accuracy is not important, and assuming distances are small (say less than 1000 miles) you can use the loxodromic distance.
For this, compute the difference of latitutes (dlat) and difference of longitudes (dlon). If there were any chance (unlikely) that you're crossing meridian 180º, take modulo 360º to ensure the difference of longitudes is between -180º and 180º. Also compute average latitude (alat).
Then compute:
distance= 60*sqrt(dlat^2 + (dlon*cos(alat))^2)
This distance is in nautical miles. Apply conversions as needed.
EXPLANATION: This takes advantage of the fact that one nautical mile is, by definition, always equal to one minute-arc of latitude. The cosine corresponds to the fact that meridians get closer to each other as they approach the poles. The rest is just application of Pythagoras theorem -- which requires that the relevant portion of the globe be flat, which is of course only a good approximation for small distances.
It all depends on what the distance is and what accuracy you require. Calculations based on "Earth locally flat" model will not provide great results for long distances but for short distance they may be ok. Models assuming Earth is a perfect sphere (e.g. Haversine formula) give better accuracy but they still do not produce geodesic grade results.
See Geodesics on an ellipsoid for more details.
One of the high accuracy (fraction of a mm) solutions is known as Vincenty's formulae. For my Excel VBA implementation look here https://github.com/tdjastrzebski/Vincenty-Excel