The Second Book of the CrossStaffChap. VI

Centesms  Minutes,  Leagues.  
If 100  do equal  60  and  20, 
Then 50  shall equal  30  and  10, 
And 5  be equal  3  and  1. 
And so the former example of 82 feet in 15 seconds, having first found that the hours way is about 21120 feet.
If I extend the Compasses from 352000, unto 21120, as before, I shall find the same extent to reach from 100 unto 6, as before, which shews that the hours way required is 6 Cent. such as 100 do make a degree, and 5 do make an ordinary league.
This might also be done at one operation. For upon these suppositions, divide 44 feet into 45 lengths, and set as many of them as you may conveniently betwee two marks on the ships side, and note the seconds of the time in which the ship goeth these lengths, so the proportion will hold,
The lengths divided by the time, shall give the Cent. which the ship goeth in an hour.
Suppose the distance between the two marks to be 60 lengths (which are 58 fette and 8 inches) and the time be 12 seconds: extend the Compasses from 12 unto 1, in the Line of Numbers; so the same extent will reach from 60 unto 5. Or extend them from 12 unto 60, and the same extent will reach from 1 unto 5, This shews that the ships way is according to 5 Cent. in an hours.
This may be found yet more easily, if the Logline shall be fitted to the time. As if the time be 45 seconds, the Logline may have a knot at the end of every 44 feet; then doth the ship run so many Cent. in an hours as there are knots veered out in the space of 45 seconds. If 30 seconds do seem to be a more convenient time, the Logline may have a knot at the end of every 29 feet and 4 inches; and then also the Cent. will be as many as the knots: or if the knots be made to any set number of feet, the time may be fitted unto the Distance. As if the knots be made at the end of every 24 feet, the Glass may be made 24 seconds, and somewhat more than an half of a second, and so these knots will shew the Cent. If there be 5 knots veered out in a Glass, then 5 Cent. if 6 knots, the ship goeth 6 Cent. in the space of an hour, and so the rest. For upon this supposition, the proportion between the time and the feet will be as 45 unto 44. But according to the common supposition, it should seem to be 45 unto 37 1/1, or in lesser terms, as 6 unto 5.
Those which are, upon the place may make proof of both, and follow that which agrees with their experience.
Extend the Compasses from the Sine of 90 gr. unto the Sine of the Complements of the Latitude; the same extent shall reach in the Line of Numbers, from the difference of the Longitude to the distance.
So the measure of one degree in the Equator being 100 Cent. the distance belonging to one deg. Of Longitude in the Latitude of 51 gr. 30 m. will be found about 62 Cent. and 1/4.
Or if the measure of a degree be 60 miles, the distance will be found about 37 miles and 1/3. If the measure be 20 Leagues, then almost 12 Leagues and 1/2 If the measure be 17 1/2 as in the Spanish Charts, then somewhat less than 11 Leagues sailing upon this parallel, will give an alteration of one degree of Longitude.
If the distance be given in Leagues or Miles, reduce them into Centesms, then will the proportion hold.
Extend the Compasses from the Sine of the Compliment of the Latitude, to the Sine of 90 gr. the same extent will reach in the Line of Numbers from the distance to the difference of Longitude.
So the distance upon a course of East or West, in the latitude of 51 gr. 30 m. being 100 Cent. the difference in Longitude will be found 1,60, which make one degree and 60 Cent. or 1 gr. 36 m.
Or if it be 60 miles, the difference of Longitude will be 96, which also make 1 gr. 36 m. as before.
Extend the Compasses in the Line of Numbers from the difference of Latitudes to the difference of Longitudes; the same extent will give the distance from the Tangent of 45 gr. unto the Tangent of the Rumb, according to the Projection of the Common Seachart.
So the latitude of the first place being 50 gr. the latitude of the second 52 gr. 30 m. and the difference of Longitude 6 gr. the Rumb will be found to be about 67 gr. 23 m. which is near the inclination of the sixth Rumb to the meridian. But this Rumb so found is always greater than it should be, and therefore to be limited; which may be done sufficiently for Seamans use, after this manner:
Extend the Compasses either from the Sine of 90 gr. unto the Sine of the Compliment of the middle Latitude, the same extent will reach from the Tangent of the Rumb before found, unto the Tangent of the Rumb limited.
Or else extend them from the Sine of 90 gr. unto the Tangent of the Rumb before found; the same extent will reach from the Sine of the Complement of the middle Latitude, unto the Tangent of the Rumb limited.
So the middle latitude between 50 gr. and 52 gr. 30 m. being 51 gr. 15 m., and the Rumb before founds 67 gr. 23 m. the Rumb limited will be found to be about 56 gr. 20 m. which is but 5 m. more than the inclination of the fifth Rumb to the Meridian.
If any please to work by the Canon, he may joyn both these in one operation.
2. This Rumb may be found by the help of the Meridian Line upon the Staff. For if I take the difference of latitude out of the Meridian Line from 50 gr. 30 m. and measure it in his Equinoctial, or at the beginning of the Meridian Line, I shall find it there to be equal to 4 gr. which may be called the difference of the Latitude inlarged. Wherefore I work as if the difference of Latitude were 4 gr.
And extend the Compasses in the Line of Numbers from 4 unto 6: so shall I find the same extent to reach from the Tangent of 45 gr. unto the Tangent of 56 gr. 20 m. and this is the inclination of the Rumb required.
Extend the Compasses from the Sine of the Complement of the Rumb, unto the Sine of 90 gr.: the same extent in the Line of Numbers shall reach from the Difference of latitude unto the distance upon the Rumb.
So the latitude of the first place being 50 gr. the latitude of the second 52 gr. 30 m. and the Rumb the fifth from the Meridian. If I extend the Compasses from 33 gr. 45 m. unto the Sine of 90 gr. I shall find the same extent in the Line of Numbers to reach from 2 gr. 50 Cent. to 4 gr. 50 Cent. and such is the distance required.
Extend the Compasses in the Line of Numbers from the distance unto the differnec of Latitudes; the same extent will reach in the Line of Sines from 90 gr. unto the Complement of the Rumb.
So the one place being in the Latitude of 50 gr. the other in the Latitude of 52 gr. 30 m. and the distance between them 4 gr. 50 Cent. If I extend the Compasses from 4,40 unto 2,50 in the Line of Numbers, I shall find the same extent to reach from the Sine of 90 gr. unto the Complement of 56 gr. 15 m. and such is the inclination of the Rumb required.
Extend the Compasses in the Line of Sines, from 90 gr. unto the Compliment of the Rumb, the same extent in the Line of Numbers, will reach from the distance, unto the difference of Latitudes.
So the lesser Latitude being 50 gr. and the distance 4 gr. 50 Cent. Upon the fifth Rumb from the Meridian: If I extend the Compasses from the Sine of 90 gr. to 33 gr. 45 m. I shall find the same extent to reach from 4,50 in the Line of Numbers unto 2,50; and therefore the second latitude to be 52 gr. 30 m.
Extend the Compasses from the Tangent of 45 gr. unto the Tangent of the Rumb; the same extent will reach in the Line of Numbers from the difference of Latitude unto the difference of Longitude, according to the Projjection of the Common Seachart.
So the first Latitude being 50 gr. and the second 52 gr. 30 m. and the Rumb the fifth from the Meridian: if I extend the Compasses from the Tangent of 45 gr. unto 56 gr. 15 m. I shall find the same extent to reach from 2,50 in the Line of Numbers to be about 3,75, which make 3 gr. 45 m. But this difference of Longitude so found, is always lesser than it should be, and therefore to be enlarged, which may be done sufficiently for Seamens use after this manner:
Extend the Compasses from the Sine of the Compliment of the middle latitude, unto the Sine of 90 gr. the same will reach in the Line of Numbers from the difference of Longitude before found, unto the difference of Longitude enlarged.
So the middle latitude in this example being 51 gr. 15 m. and the difference of Longitude before found, 3 gr. 75 Cent. the difference of the Longitude inlarged will be found about 5 gr. 99 Cent. which are near 6 gr.
If any please to work by the Canon, he may joyn both these in one operation.
2. This difference of Longitude may be found by help of the Meridian Line upon the Staff. For if I take the proper difference of latitude out of the Meridian Line, and measure it in his Equinoctial, or at the beginning of the Meridian Line, I shall find the Latitude inlarged to be equal to four of those degrees.
Wherefore having extended the Compasses, as before, from the Tangent of 45 gr. unto the Tangent of 56 gr. 15 m. the same extent will reach from 400 in the Line of Numbers, unto 5,99, which shews the difference of Longitude to be abour 5 gr. 99 Cent, or about half a minute short of 6 degrees.
Take the proper difference of Latitudes out of the Meridian Line of the Chart, and measure it in his Equinoctial, or one of the Parallels, and it will there give the difference of Latitude enlarged.
Then extend the Compasses from the Sine of the Compliment of the Rumb unto the Sine of 90 gr. the same extent will reach in the Line of Numbers, from the latitude enlarged, unto the distance required. Or extend them from the Compliment of the Rumb to the latitude inlarged, the same extent will reach from 90 gr. unto the distance.
For example, Let the place given be A, in the Latitude of 50 gr. D, in the latitude of 52 gr. 30 m. AM the difference of Latitudes, and the Rumb MAD the fifth from the Meridian. First, I take out AM, the difference of latitudes, and measure it in AE, one of the parallels of the Equinoctial; I find it to be very near 4 gr. this ist the difference of latitudes inlarged. Then if I extend the Compasses from the Sine of 33 gr. 45 m. the Compliment of the fifth Rumb, unto the Sine 90 gr. I shall find the same extentto reach in the Line of Numbers, from 400 unto 7,20, And this is the distance belonging to the Chart. Wherefore I take out these 7 gr. 20 Cent. out of the Scale of the parallel AE, and prick it down upon the Rumb from A unto D, where it meeteth with the parallel of the second Latitude. Lastlly, I measure it in the Meridian Line, setting one foot of the Compasses as much below the lesser Latitude, as the other above the greater latitude, and find it to be 4 gr. 50 Cent. which is the same distance that I found before in the 5 Prop.
The way of the Ship may be known as in the first Prop. The Angles may be observed either by the Staff, or by a Needle set on the Staff. For example, suppose that being at A, I had sight of the Land at B, the Ship going East Northeast from A toward C, and the Angle of the Ships Position BAC being 43 gr. 20 m. and after that the Ship had made 10 Cent. or two Leagues of way from A unto D, I observed again, and found the second Angle of the Ships Position BDC to be 58 deg. Or the inward Angle BDA, to be 112 deg. Then may I find the third Angle ABD, to be 14 deg. 40 m. either by Subtraction, or by Complement unto 180 gr.
In this and like cases, I have a right Line Triangle, in which there is one side and three Angles known, and it is required to find the other twi sides, and the Canon for it is this:
Wherefore I extend the Compasses from 14 gr. 40 m. in the Sines, to 10 in the Line of Numbers, and this extent doth reach from 58 gr. to 33 1/2 , and such is the distance between A and B, and it reacheth from 43 gr. 20 m. unto 27 in the Line of Numbers; and such is the distance from D to B. These two distances being known, I may set out the land upon the Chart. For having set down the way of the Ship, from A to D, by that which I shewed before in the use of the Meridian Line, I may by the same reason set off the distance A B and D B, which meeting in the Point B, shall there resemble the Land required.
If it may be conveniently, let the Angle of Position be observed at such time as the Ship cometh to be right over against one of the places. As if the places be East and West, seek to bring one of them South or North of you, and then observe the Angle of Position, so shall you have a right Line Triangle, with one side and three Angles, whereby to find the two other sides. First, you have the Angle or Position at the Ship, then a right Angle at the place that is over against you, and the third Angle at the other place is the Complement to the Angle of Position. Wherefore,
So the places being 15 Cent. or three Leagues one from the other, and the Angle of Position 29 gr. the nearer distance will be found about 12 Cent. and the further distance about 31 Cent.
Or howsoever the Angle of Position were observed, the distance between Ship and Land may be found generally as in this example.
Suppose A and D were two head Lands known to be East Northeast, and West Southwest, 10 Cent. or two Leagues one from the other; and the Ship being at B, I observe the Angle of the Ships Position DBA, and found it to be 14 gr. 40 m. and that D did bear 9 gr. 30 m. and A 24 gr. 10 m. from the Meridian BS, and this example would be like the former. For if the Angle SBD be 9 gr. 30 m. from the South to the Westward, then shall NDB be 9 gr. 30 m. from the North to the Eastward. Take these 9 gr. 30 m. out of the Angle NDE, which is 67 gr. 30 m. because the two head Lands lie East Northeast, and there will remain 58 gr. for the Angle BDE, and the inward Angle BDA out of 180 gr. Take these two Angles ABD and BDA out of 180 gr. and there will remain 43 gr. 20 m. for the third Angle BAD. Wherefore here also are three Angles and one side, by which I may find the two other sides, as in the last Prop.
These Propositions thus wrought by the Staff, are such as I thought to be useful for Seamen, and those that are skilful may apply the example to many others. Those that begin, and are willing to practice, may busie themselves with this which followeth.
Suppose four Ports L, N, O, P, of which L is in the latitude of 50 Gr. N is North from L 200 Leagues or 1000 Cent. O West from L 1000 Cent. and P West from N 1000 Cent. so that L and O will be in the same latitude of 50 gr. N N and P both in the Latitude of 60 gr. Then let two Ships depart from L, the one to touch at O, the other at N, and then both to meet at P, there to Lade, and from thence to return the nearest way unto L. Here many questions may be proposed:
These questions well considered, and either resolved by the Staff, or pricked down on the Chart, and compared with the Globe and the common Seachart, shall give some light to the direction of a course, and reduction of places to their due Latitude, which are now fully distorted in the common Seacharts.
Here follows all the usual Problems of sailing, according to Mercator, which are resolved Arithmetically by the Table of Logarithm Tangents, without the Table of Meridional parts, and may also be performed Geometrically, by the Tangent Line upon the Crossstaff if it be large.
First, we are to know that the Logarithm Tangents from 45 gr. 00 m. upward, do increase in the same maner, that the Secants added together do, if we account every half degree above 45 gr. 00 m. to be one whole degree of Mercators Meridional Line: and so the Table of Logarithm Tangents, is a Table of Meridional parts, to every two minutes of the Meridian Line, leaving out the Radius in every Line.
The manner of making use of it thus, (as it shall more plainly appear in the Examples of the following Problems) because the Tables begin at 45 gr. 00 m. and that every 30 m. is for a whole gegree, when one, or both Latitudes are given in any questions, take 1/2 of each Latitude, and add 45 gr. 00 m. to each of them, and take the Tangent of the sum of each, for equal parts of the Latitudes given (neglecting the Radius as before said) then subtract the lesser sum of equal parts from the greater, and the remainder divide by the Tangent of 45, 30, the Radius neglected, the Quotient shall be equal or Equinoctial gegrees contained between the two Latitudes, or else multiply the foresaid remainder by 10, and divide it by the half of the foresaid Tangent of 45 gr. 30 m. and the Quotient shall be equal or Equinoctial Leagues contained between the two Latitudes.
Example.
Let one Latitude be 45 gr. 39 m. the 1/2 is 22 gr. 45 m. unto which add 45 gr. 00 m. the sum is 67 gr. 45 m. the Tangent above the Radius is 3881591. Let the other Latitude be 40 gr. 00 m. the 1/2 is 20 gr. 00 m. unto which add 45 gr. 00 m. the sum is 65 gr. 00 m. the Tangent above the Radius is 3313275, which subtracted from the former, the remainder is 56831: which being divided by 75803 the Tangent of 45 gr. 30 m. above the Radius, the Quotient is 7 gr. 497 parts, the Equinoctial degree contained between the two latitudes, or else multiply the remainder or difference 568316 by 10 and divide it by 37901, the 1/2 Tangent of 45 gr. 30 m. above the Radius, and the Quotient is 149 Lea 94 parts, the equal Equinoctial Leagues contained between the two Latitudes, and the like of any other.
Admit a Ship sail SSE 3/4 E 128 Leagues from the Latitude 45 gr. 30 m. North Latitude, that is from A to E, according to the plain Seachart, I demand the true place or point that the Ship is at, according to Mercators Chart.
Before this question can be resolved, we must find ahat Latitude the Ship is in, which is thus found:
As the Radius is to the Sine of the Complement of the course 50 deg. 04 m. So is EA the distance upon the course 128, to AD the true difference of Latitude in Leagues, which is 110. This being converted into deg. And min. is 5 deg. 30 m. and because the latitude decreaseth, or the Pole is depresses, we subtract it from 45 d. 30 m. and the remainder is 40 deg. 00 M. the Latitude the Ship is in, that is at E, according to the plain Seachart, or at C according to Mercator: but before we can find the point C, we must find the distance of the point B in the Meridian Line from A: the manner how to do it is shewed in the Example before this Problem, and it is there found to be 150 Leagues near. Now the point C, the true place of the Ship in Mercators Chart may be found two several ways.
First, As DA the true distance of Latitude 110, is to AE the true distance run upon the course, so is BA the difference of Latitude enlarged 150, to AC 174 5/10 the enlarges distance, which being laid off upon the Line of the course, gives the point C, the true place of the Ship in Mercators Chart. Here we may take notice, that the true point of any place, according to the plain Seachart, or according to Mercators Chart, is always upon one and the same right Line of the course.
Secondly, As the Radius is to the Tangent of the Course 30 d. 56 m. so is AB the difference of Latitude enlarged 150, to BC 90 the difference of Longitude, which being laid off upon the perpendicular BC, gives the point C, the true place of the Ship in Mercators Chart.
Suppose a Ship to sail SSE 1/4 E from the Latitude of 45 deg. 30 m. until it be in the Latitude of 40 deg. 00 m. that is from A to E, according to the plain Seachart, or from A to C, according to Mercators Chart.
First, we must find the difference of latitude enlarged, as is before directetd, which is AB 150 Leagues.
Supose a Ship to sail 128 Leagues, between South and East, from A in the Latitude of 45 deg. 30 m. and at the ende of her distance, it be in the Latitude of 40 deg. 00 m.
First, find the difference of Latitude enlarged, as before directed, which is AB 150.
That Problem is chiefly useful for the Navigator, when he hath cast up his traverse. Admit a Ship to sail upon the Southeast quarter of the Compass, from latitude 45 deg. 30 m. unto Latitude 40 deg. 00 m. and the departure from the Meridian it went from, be 65 8/10 Leagues.
First, find the difference of latitude enlarged, as before directed 150 Leagues.
Admit a Ship to be at A in North Latitude 45 ge. 30 m. and to sail Southeastwards, until it be at E in Latitude 40 deg. 00 m, according to the plain Chart, and the point C be the place in Mercators Chart where the Ship is, and the difference of Longitude be BC 90 Leagues.
First, find the difference of Latitude enlarges, as is before directed 150 Leagues.
Suppose a Ship to be in the Latitude of 45 deg. 30 m. North Latitude, and to sail SSE 1/4 E (until the difference of Longitude be 90 Leagues) that is from A to C, which is the point or place of the Ship in Mercators Chart.
Although I have set down but these proportions and the answers to each question, they may all be calculated by the Canon, and the Chiliad of Logarithms in this Book.
© Rainer Stumpe URL: http://www.rainerstumpe.de 