896 Words4 Pages

Lab Report 1
Logarithmic Plotting
Devin Edwards
ENGR 3070L
CRN: 27194
January 17, 2018
Dr. Margraves Objective
The purpose of this experiment is to graph and look at the logarithmic plots and write corresponding exponential equations that match the “Best Fit” line of the data points.
Theory
The data in Table 1 can be represented by the exponential equation given in equation 1 below. Equation 1 is also used for Cartesian plots:
Q=KH^n (1)
On this type of plot a straight line is drawn representing the slope intercept form in equation 2: y=mx+b (2)
The variables are switched out allowing it to represent the same relationship in logarithmic form, where y is log(Q), x is log(H), n is m, and b is log(K): log(Q)=log(K)+n[log(H)] (3)*…show more content…*

In the first method two random points are chosen along the best fit line of the data. The vertical distance and horizontal distance are measure in cm. The distance in the vertical direction is divided by the distance in the horizontal direction “rise over run”. Using this particular method I calculated that the value of n is .604. The second method was also used considering that two points along the best fit line were exactly one cycle away, meaning the value of H2 was ten times larger than the value of H1. Equation 3 was used in this method. It became easier since the value of log10 is equivalent to 1. After Q1 and Q2 were chosen at the H values, the next step is to solve for log Q2/Q1. This calculated value of n came out to be 0.5057. There were also two different methods for solving for the value of K using the best fit line given in Figure 1. By finding where the first K is equal to 1 and seeing what the Q value is at this given point, K can be solved. As said above when H equals 1 the Q value is the same as the K value. The K value was calculated to be 4.7. In the second method for solving for K, use the values calculated for the variables Q, K, and n and plug into Equation 1. By doing this the calculated value was 4.645. The value of K calculated in excel was calculated to be

In the first method two random points are chosen along the best fit line of the data. The vertical distance and horizontal distance are measure in cm. The distance in the vertical direction is divided by the distance in the horizontal direction “rise over run”. Using this particular method I calculated that the value of n is .604. The second method was also used considering that two points along the best fit line were exactly one cycle away, meaning the value of H2 was ten times larger than the value of H1. Equation 3 was used in this method. It became easier since the value of log10 is equivalent to 1. After Q1 and Q2 were chosen at the H values, the next step is to solve for log Q2/Q1. This calculated value of n came out to be 0.5057. There were also two different methods for solving for the value of K using the best fit line given in Figure 1. By finding where the first K is equal to 1 and seeing what the Q value is at this given point, K can be solved. As said above when H equals 1 the Q value is the same as the K value. The K value was calculated to be 4.7. In the second method for solving for K, use the values calculated for the variables Q, K, and n and plug into Equation 1. By doing this the calculated value was 4.645. The value of K calculated in excel was calculated to be

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