the required number of miles he must travel on ENE bearing for every 10 miles he deviated. Remember, his alargar (distance from intended course) was 55 miles. So in order to return to his intended course he must travel 5.5 × 26 = 143 miles on ENE. In other words, he needs to hold his ENE bearing for 143 miles; once that distance is traveled, he should straighten his ship east, and he will be exactly back on the intended course. The sixth and final column (avanzo di ritorno) gives the length on the intended course he has made good by his return travel. This is also expressed in terms per 10 miles alargar. His alargar was 55, and his angle of return was ENE (thus q = 2), that means his avanzo di ritorno is 5.5 × 24 = 132. In other words, if everything goes right, and our mariner holds his ENE bearing for 143 miles (ritorno), then during that return, he will have covered an additional 132 miles on his intended eastward course (avanzo di ritorno). Finally, to calculate the total distance made good (total avanzo) on the eastward bearing by his whole adventure, he must add the avanzar during the deviation (83 miles) plus the avanzo di ritorno (132 miles). Thus on the whole, he has covered 83 + 132 = 215 miles on the intended course. Measuring that distance on the map from the starting point (A), the mariner can figure out his exact current position. This is the simplest use of the toleta de marteloio. It is, at root, a trigonometric table. However, it does not tackle the traverse problem in one go, like the Law of Sines, but rather splits the problem into two right-angled triangles which it proceeds to solve successively. Modern trigonometry would dispense with the step of calculating the alargar, and calculate the ritorno directly – but for that, one needs to be armed with a full sine table. The toleta is a rather simple table, easy to consult and perform calculations with, and sufficiently compact to be memorized by navigators (as Bianco recommends). Rule of three The toleta de marteloio is expressed for nice round numbers, 100 and 10. But, in practice, a ship would not usually sail 100 miles before trying to return, but some other distance, say 65 miles. To calculate this is a simple problem of