Estimating the Mass of Mount Everest
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- Nathaniel Hutchinson
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1 Estimating the Mass of Mount Everest Mathematica is a great tool for exploring geographical objects such as deep ocean trenches, high mountains or even other planets. Here we will use its rich database to estimate the approximate mass of mount Everest. However, first we need to define what constitutes mount Everest. For me the coherence of the general procedure is more important rather than the specific area we would like to include as this could be recalibrated. For the purpose of the example, let s say that mount Everest is defined from 0m altitude up to the surface layer within a circle with a radius of 5km, centered at the peak. For our estimation, we will define the mass as the product of volume times the rock density. The slight variations in the gravitational force at sea level and at the higher altitudes will be ignored. Let s see how we could set up the problem: Step 1. Get the location and altitude data everest = GeoElevationData GeoDisk Entity "Mountain","MountEverest", Quantity 5,"Kilometers" ; Let s see the output in the form of an array: everest QuantityArray Dimensions: {16, 19} Unit: Meters There are 16 measurements in longitude and 19 in latitude. What this data tells us is that there is a vertical measurement every 5km/16= 312.5m and horizontal one every 5km/19=263.2m. It is not perfectly symmetric. To represent the whole surface we will perform an interpolation using all available points. Lastly, notice the units associated with the numbers. They slow down the computation, so we will keep them in mind but discard them from the estimation procedure which is performed by the following code: everest`no`units=quantitymagnitude[%3];
2 2 everest notes.nb Putting the data in matrix form: MatrixForm everest`no`units (*will be truncated a bit because of its size*) 5855.` 5775.` 5785.` 5868.` 6166.` 6792.` 7164.` 7084.` 7040.` 7060.` 6812.` 6501.` 6406.` 6367.` 6365.` 6483.` 6683.` 6536.` 6041.` 5845.` 5839.` 5906.` 6198.` 6804.` 7100.` 7279.` 7081.` 6744.` 6601.` 6512.` 6457.` 6437.` 6443.` 6471.` 6592.` 6670.` 6069.` 5903.` 5881.` 5917.` 6012.` 6318.` 6525.` 6741.` 6933.` 6793.` 6644.` 6606.` 6624.` 6744.` 6777.` 6551.` 6438.` 6315.` 5982.` 5942.` 5969.` 6016.` 6060.` 6131.` 6231.` 6479.` 6830.` 7108.` 7049.` 6985.` 7079.` 7181.` 6830.` 6357.` 6013.` 5806.` 6002.` 6226.` 6502.` 6317.` 6252.` 6322.` 6417.` 6681.` 7068.` 7411.` 7605.` 7610.` 7595.` 7424.` 6845.` 6301.` 5956.` 5873.` 6065.` 6601.` 6917.` 6831.` 6792.` 6906.` 7015.` 7205.` 7440.` 7729.` 8027.` 8045.` 7696.` 7340.` 6932.` 6503.` 6261.` 6119.` 6034.` 6592.` 7144.` 7204.` 7298.` 7359.` 7467.` 7668.` 7876.` 8132.` 8316.` 7854.` 7385.` 7006.` 6501.` 5964.` 5786.` 5773.` 6032.` 6352.` 6804.` 6916.` 6997.` 7140.` 7498.` 7925.` 8304.` 8536.` 8186.` 7650.` 7266.` 6892.` 6460.` 5950.` 5663.` 5510.` 6098.` 6183.` 6376.` 6478.` 6575.` 6734.` 7072.` 7551.` 8100.` 8499.` 8177.` 7697.` 7277.` 6874.` 6510.` 6048.` 5595.` 5471.` 6685.` 6385.` 6337.` 6359.` 6431.` 6510.` 6736.` 7150.` 7634.` 8088.` 8064.` 7656.` 7171.` 6656.` 6088.` 5769.` 5552.` 5467.` 7242.` 7122.` 7087.` 6811.` 6607.` 6589.` 6654.` 6888.` 7333.` 7744.` 7913.` 7571.` 7063.` 6522.` 5960.` 5620.` 5523.` 5486.` 6970.` 7442.` 7555.` 7221.` 6953.` 6791.` 6773.` 6945.` 7329.` 7723.` 8027.` 7726.` 7164.` 6518.` 5850.` 5633.` 5565.` 5519.` 6197.` 6868.` 7290.` 7541.` 7419.` 7262.` 7198.` 7267.` 7547.` 7915.` 8267.` 8182.` 7640.` 7043.` 6435.` 6083.` 6068.` 6115.` 5772.` 6303.` 6732.` 6862.` 7220.` 7243.` 7205.` 7323.` 7532.` 7601.` 7597.` 7704.` 8007.` 7838.` 7467.` 7259.` 7232.` 7002.` 5733.` 5991.` 6147.` 6204.` 6518.` 6654.` 6418.` 6317.` 6379.` 6463.` 6613.` 6887.` 7118.` 7408.` 7421.` 7246.` 7166.` 7026.` 5614.` 5542.` 5467.` 5541.` 5703.` 6076.` 6026.` 5755.` 5594.` 5642.` 5890.` 6252.` 6591.` 6862.` 6975.` 6754.` 6611.` 6629.` I like to think of such matrices in terms of contour maps, as the entry indices signify the spatial coordinates, while the entry itself is the height/output of some function. Before creating contour maps, let s use a discrete plot before doing any interpolations. ListPointPlot3D everest`no`units, Filling Bottom, ColorFunction "TemperatureMap", PlotTheme "Marketing", PlotLegends Automatic, Axes True, AxesLabel Style "longtitude",13, Black,"latitude",Style "height",13, Black, AxesStyle Directive[Black, 10]
3 everest notes.nb 3 Except for the height, the spatial coordinates are represented in a big chunks of meters, it is not normalized, but this should not worry us for the moment. ListPlot3D everest`no`units, Mesh Automatic, MeshFunctions {#3&}, ColorFunction ColorData["BeachColors"], PlotTheme "Marketing", PlotLegends Automatic, Axes True, AxesLabel Style "longtitude",13, Black,Style "latitude",13, Black, Style "height",13, Black, AxesStyle Directive[Black, 10]
4 4 everest notes.nb Let s do a contour map as well ListContourPlot everest`no`units, Mesh Automatic, MeshFunctions {#3&}, ColorFunction->"TemperatureMap", PlotTheme "Marketing", PlotLegends Automatic, FrameStyle Directive Black,Thick So, we have the graphical representation of the mountain region. Now we want to find some multivariable function that will fit the real data, and then use double integrals to compute the volume. interpolation1 = ListInterpolation everest`no`units InterpolatingFunction Domain: {{1., 16.}, {1., 19.}} Output: scalar This is not in the appropriate format to be plugged into a double integral yet, as we haven t given names to the spatial coordinates. Let us rewrite it and use x and y as variables for the integration. function = interpolation1[x,y] InterpolatingFunction Domain: {{1., 16.}, {1., 19.}} Output: scalar [x, y]
5 everest notes.nb 5 Now we have a general functional representation of the mountain terrain and can integrate it. The boundaries are defined in the rescaled units set by Mathematica. volume = NIntegrate function,{x,1,16},{y,1,19} This answer is nonsensical until we recalibrate it by multiplying by axis values by their values in terms of meters. Recall that 1x unit = 312.5m and 1y unit=263.2m I will just take the rough average of the two at about 290m. volume2 = 290*volume (* in m 3 cubic meters *) Lastly, in order to get the mass, we need to use the rock density as well. The average rock density in terms of cubic meters is 2650 kg m 3. rock`density = 2650; mass =rock`density * volume This is the approximate mass of the region (in kg) we defined as Everest. The result is more dependent on the definition used, but the approach has the benefit that applies the integration on an interpolated shape very close to the original, especially in comparison to perfect geometric figures as cones.
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