Interperating a graph with logarithmic scales

  • Thread starter ehilge
  • Start date
  • #1
ehilge
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Homework Statement


see attachment


Homework Equations


log10x=y
10y=x


The Attempt at a Solution


Since I'm rather confused on reading a logarithmic scale, I thought I'd post this to see if I'm doing this right and/or if there's a better way to do the problem.

a) T=500 C
using the # of intervals we can say that if d=.01, x=.5 intervals so...
log(t)=.5
t=3.16 min

similarly for d=.1
log(t)=3.5 intervals
t=3162

change in time = 3162-3.16 = 3158.8min

b) same process but use x=-.5 intervals and x=2.1

If this method does work, is there any way to interpolate values so I can be more exact? or is there a better way to do the problem in general?
thanks!
 

Attachments

  • logarithmic scale.jpg
    logarithmic scale.jpg
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Answers and Replies

  • #2
36,203
8,189
I think you're on the right track, but maybe just a little off. For the 500 degree track (part a), the starting time is about 3.2 min, but the ending time looks to me like about 3.6, which corresponds to about 4000 min. Subtracting gives a duration that's again close to 4000 min. Unless you can really measure the position of the points very precisely, it doesn't make sense to end up with answers with more than one or two places of precision.
 
  • #3
ehilge
163
0
yah, I was worried about being able to get a precise enough answer to matter. I think later I'll find a ruler and figure out more exact data points
thanks!
 
  • #4
36,203
8,189
Yes, a ruler would help you get more precise answers.
 

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