# NavList:

## A Community Devoted to the Preservation and Practice of Celestial Navigation and Other Methods of Traditional Wayfinding

**Re: Using logs.**

**From:**Frank Reed

**Date:**2008 Apr 23, 21:20 -0400

coralline algae writes: "at page 38 is the start of log tables for sin, cos and tan and what was done was to add the number 10 to the log10 result so no logarithms are negative. I havent actually tried to use these tables as yet but I am guessing that this is another workaround as discussed in this thread?" That solves a different problem, also important. Let's consider a simple trig calculation. We want to work out the product x = tan(a)*sin(b) where a=-60 degrees and b=+15 degrees. Work it out with modern tools, a calculator or in software, and you will find that x is -0.448288. But let's do it with logarithms... The problem x=tan(a)*sin(b) is equivalent to x=invlog[logtan(a)+logsin(b)] where the function "invlog(z)" means 'find the number whose logarithm is equal to z' and is equivalent to 10^z. First, we look up the logtangent of 60 degrees (ignoring the sign of the argument). That's 0.2385607. Next we look up the logsine of 15 degrees. That should be -0.5870038. Now here's the trick. Instead of dealing with subtraction, which is generally a little more error-prone than addition, we just add ten to this negative logarithm making it 9.4129962. That is, the folks who constructed the tables add this 10 so that we never see the negative value of the logarithm at all. If you think about it, this is equivalent to multiplying the original problem by ten billion (10 to the tenth power). We can get away with this trick as long a number this large would never appear in any of the calculations we might do. Then, we add the logarithms. The result is 9.6515569. If the result had been greater than ten, we would "drop the tens" (a common instruction in old navigation manuals) which we would today refer to as taking the result "mod 10". We would then "seek" this logarithm in a table of standard logarithms and read out the result: 0.448288. And finally, we have to remember that negative sign that we ignored at the very beginning. So far we've done the calculation as if the arguments were both positive. But since the argument of the tangent was actually -60 degrees, the factor itself would be negative and therefore, so is the final answer: -0.448288. You can see that fussing over signs could get to be fairly complicated. The instructions for solving this problem in a traditional source might read something like this: "if the argument of the tangent is negative or the argument of the sine is negative (but not both negative), then the final result should be marked negative, otherwise positive". These conditional statements could get to be difficult to follow so practical mathematicians always looked for solutions to problems with a minimal number of possible signs. -FER --~--~---------~--~----~------------~-------~--~----~ Navigation List archive: www.fer3.com/arc To post, email NavList@fer3.com To unsubscribe, email NavList-unsubscribe@fer3.com -~----------~----~----~----~------~----~------~--~---