Math Nerds

[HoustonHorn]

Need help.

I have two x,y coordinates

One is .275 hours on the x axis and 4.45 ft on the y axis (.275,4.45)

The second point is (2.31 hrs,15 ft)

I am looking for (x,8.27).

Easy enough on a straight line, but I have limited data points (2) when I know in actuality, the calculation it isn't linear. There is definitely a somewhat exponential progression as the elevation (y-coordinate) gets deeper, the time (x-coordinate) extends more.

Basically, its a drilling operation, so the deeper you go, the tougher it gets, so I can't use a linear extrapolation, but without an additional data point I can't easily determine what sort of curve is reasonable.

This is an estimate, so it doesn't have to be perfect, but I would like to come up with something in the middle of exponential and linear so I can see what my margin of error is between the two and use that for the estimate.

Is there a way to determine the difference in the rate at point 1 and the rate at point 2 to determine what the "curve," is with only two data points?

Any thoughts? Thanks in advance.

 

[HoustonHorn93]

You lost me at:

In reply to:

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[Doughbroz]

That question is so simple, I'm gonna let my limo driver answer it.
Jerry Clower

 

[Math Mudrat]

Easy way to do this in Excel:

Put your points on a graph and tell it to "Add Trendline." Do a linear fit and write down the equation. Then do an exponential fit and write down the equation. The difference in the values that the two methods produce gives you some idea of a systematic error.

 

[AstroVol]

Remember that (0,0) is also 3rd data point that will help with the curve fitting.

 

[HoustonHorn]

Thanks for the help. I've never done a trendline in Excel, but I will give it a shot.

 

[HoustonHorn]

Thanks for the help, that worked and gave me a conservative estimate.

 

[Steel Shank]

I'm not even going to dignify this elementary question with a response.

 

[BadgerinATX]

For your linear question, you can get the slope m = (y2 - y1)/(x2 - x1). Then you can use the point-slope formula to find the x value corresponding to y = 8.27. The point-slope formula is y - y1 = m(x - x1). You can use any point on the line - in your case, either of the two points you know. You know x1 and y1. You know y = 8.27 and m = whatever you get. Now you just need to solve for x w/ some algebra.

Your second question is more involved. If you only know two points, the best and only fit you can obtain is linear. It sounds like your data would be closer to t^n, where 0<n<1, or a logarithmic relationship.

Would it be safe to say that depth(t=0) = 0? If so, you now have three data points.