tag:blogger.com,1999:blog-4237678261594285909.post4293119881873167685..comments2023-10-15T09:52:03.198-04:00Comments on Sailboat Instruments: Leeway calibration (Part 3)Unknownnoreply@blogger.comBlogger15125tag:blogger.com,1999:blog-4237678261594285909.post-74465636432315746742012-04-10T13:25:08.279-04:002012-04-10T13:25:08.279-04:00Thanks Merlin, will give that a try.
MartinThanks Merlin, will give that a try.<br /><br />MartinMartinnoreply@blogger.comtag:blogger.com,1999:blog-4237678261594285909.post-65890931224660622642012-04-07T09:57:56.218-04:002012-04-07T09:57:56.218-04:00To complete my last comment, here how you can chec...To complete my last comment, here how you can check your compass calibration. If your system can log continuously the heading and the COG, you can check the compass deviation with the procedure described in this post:<br />http://sailboatinstruments.blogspot.ca/2011/01/gyro-compass-calibration.html<br /><br />You can also check the tilt compensation correction at dock, by heeling the boat with an halyard while keeping it parallel to the dock, and see if the heading remains constant.Merlinhttps://www.blogger.com/profile/00901116173524809046noreply@blogger.comtag:blogger.com,1999:blog-4237678261594285909.post-22159571057804119152012-04-06T18:21:21.835-04:002012-04-06T18:21:21.835-04:00In a document that I refer to elsewhere in this bl...In a document that I refer to elsewhere in this blog, Al Gentry writes : "Values for the constant K vary between 9 and about 16, depending on the winward efficiency of the boat". It is possible that modern boats are better than that, but your calculated leeway value appears indeed suspiciously low. I have had bad experiences with calibration of commercial fluxgate compass. Errors of several degrees are not uncommon. You may want to compare your heading and COG in waters with really negligible currents.Merlinhttps://www.blogger.com/profile/00901116173524809046noreply@blogger.comtag:blogger.com,1999:blog-4237678261594285909.post-44843248521123194152012-04-06T17:47:15.534-04:002012-04-06T17:47:15.534-04:00Thats great.
The magnetic compass is a B&G Ha...Thats great. <br />The magnetic compass is a B&G Halcyon 2000.<br /><br />Why is the K value so low. Ain't it suppose to be higher?<br /><br />Thanks for your help.Martinnoreply@blogger.comtag:blogger.com,1999:blog-4237678261594285909.post-18066960958951888882012-04-06T17:29:19.802-04:002012-04-06T17:29:19.802-04:00I reproduce exactly your results. The important th...I reproduce exactly your results. The important thing is that the headings and COGs are consistent (both magnetic, or both true).<br /><br />Be aware that the accuracy of this result depends mostly on the accuracy of the magnetic compass.Merlinhttps://www.blogger.com/profile/00901116173524809046noreply@blogger.comtag:blogger.com,1999:blog-4237678261594285909.post-55825504678798731182012-04-06T17:16:09.299-04:002012-04-06T17:16:09.299-04:00Hi Merlin
In Expedition its says Cog °M and Hdg °...Hi Merlin<br /><br />In Expedition its says Cog °M and Hdg °M. <br />Guess its magnetic then?<br /><br />The B&G system is also set in magnetic mode from Expedition.Martinnoreply@blogger.comtag:blogger.com,1999:blog-4237678261594285909.post-32342407587511571572012-04-06T13:33:00.280-04:002012-04-06T13:33:00.280-04:00Hi
First off, thanks for an excellent site
I hav...Hi<br /><br />First off, thanks for an excellent site<br /><br />I have tried to calculate leeway for my boat.<br />Could somone just verify that my numbers are correct?<br /><br /><br />These are the numbers(Expedition Stripchart also gives some numbers inside brackets when using wands, anyone know what they are?<br /><br />Upwind<br /><br />Sog 6,085 [6,074] sd 0,192 <br />Cog 125,1 [127,7T] sd 1,3<br />Heading 131<br />Heel 10,23<br />Bsp 5,904<br /><br />Dead downwind<br /><br />Sog 4,113 [3,957] sd 0,177<br />Cog 8,2 [8,6T] sd 1,8<br />Heading 8,5<br />Bsp 3,180<br /><br />Calculated leeway: 1,30548054745469<br />Calculated K value: 4,44823239750014<br /><br />Is these numbers correct?Martinnoreply@blogger.comtag:blogger.com,1999:blog-4237678261594285909.post-14476526720353645262012-03-25T21:18:22.896-04:002012-03-25T21:18:22.896-04:00No, my mistake, the 3.72 is correct. The correctio...No, my mistake, the 3.72 is correct. The correction has been made in the table. I just realize that the 3rd comment in this post had already flagged this.Merlinhttps://www.blogger.com/profile/00901116173524809046noreply@blogger.comtag:blogger.com,1999:blog-4237678261594285909.post-78120195585309576572012-03-25T18:19:04.493-04:002012-03-25T18:19:04.493-04:00Hello,
For:
// Calculate VMG (velocity made good)...Hello,<br /><br />For:<br />// Calculate VMG (velocity made good)<br />vmg = stw * cos((-twa + leeway) * DEG_TO_RAD);<br /><br />6.05 * cos( ( 45 + 7 ) * ( 3.14 / 180 ) )<br /><br />I am getting 3.72, not 4.24. I got the calculation above from your post on May 26,2011.<br /><br />Am I missing something?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-4237678261594285909.post-14542571119812618752011-02-20T16:31:11.899-05:002011-02-20T16:31:11.899-05:00You are right: I use the TWA compared to heading. ...You are right: I use the TWA compared to heading. I should call it the “displayed TWA” that will be show on a graphic instrument with the boat pointing up. I agree that it is not the true TWA compared to course.<br /><br />Now the simple answer to the 2.57 deg change of (displayed) TWA for system C to B is that system B is wrong and that we are just measuring the resulting error that it introduces by the incomplete calculations.<br /><br />A more analytical answer is that when the tip of the blue arrow passes form the tip of the red arrow to the tip of the broken arrow, its angle changes by 2.57 deg.<br /><br />If I had adopted to use the ‘true’ TWA compared to course, the TWA would have changed from 45 +7=52 to 42.43 deg, because for system B, the course is the heading.Merlinhttps://www.blogger.com/profile/00901116173524809046noreply@blogger.comtag:blogger.com,1999:blog-4237678261594285909.post-8146208320758042732011-02-20T06:03:19.021-05:002011-02-20T06:03:19.021-05:00First a compliment for your math and explanation, ...First a compliment for your math and explanation, it just too much sometimes for my brain.<br /><br />when using the spreadsheets (and doing some tests in my navigation program) I see that they just add the leeway to the TWA angle. So this is TWA compared to course, and you are calculating TWA compared to heading?<br /><br />So when I start with a TWA of 52, AWS of 12 and STW of 6.05 I get a AWA of 35,13<br /><br />Then they subtract the 7 degrees of leeway and I end up with 28.13 (close to your 28.17)<br /><br />The strange thing about your calculation is that with 7 degrees of leeway, the AWA changes just 2.57 degrees from system C to system B.<br />Can you explain that?Erikhttps://www.blogger.com/profile/10896128673707323955noreply@blogger.comtag:blogger.com,1999:blog-4237678261594285909.post-33990153061364534722011-02-20T01:12:23.278-05:002011-02-20T01:12:23.278-05:00First a clarification on measured boat speed (MBS)...First a clarification on measured boat speed (MBS) vs speed through water (STW). When using a paddlewheel as speed sensor, MBS and STW are different when there is leeway. The paddlewheel is installed parallel to the centerline of the boat, so that it measures the component of the water flow parallel to the centerline of the boat. When there is leeway, the water under the boat does not flow parallel to the centerline, but a little sideways (by the leeway angle). The paddlewheel has been calibrated in a situation where there was no leeway. So in the diagram above, what the paddlewheel is measuring is the broken arrow (MBS). But the real path though water is the red arrow. One of the tasks of a system that corrects for leeway is to calculate and display the real speed through water STW = MBS/cos(leeway). This is not a big correction (0.05 knot), but it is there.<br /><br />Now an explanation on how this example has been developed. The third column is calculated first, to represent correctly the diagram with the following imposed conditions: TWA = 45.0 deg, TWS = 12.0 knots, MBS = 6.0 knots, heel angle = 20.0 deg, K = 12.6, heading = 0 deg, no upwash, no wind shear. From these inputs, the following values are calculated: leeway = 7.0 deg, STW = 6.05 knots, AWS = 16.43 knots, AWA = -28.17 deg. Note that for these conditions, the wind vane will measure an uncorrected AWA of -26.71 deg, but after the correction for heel, system C will report the correct value of 28.17. System C will be able to retrieve the correct true wind speed conditions (the blue arrow), because it can measure correctly the green and red arrows. System C will also calculate correctly the VMG and the wind direction.<br /><br />Now suppose that for these same conditions, the instrumentation system makes no correction for leeway (that is system B, the second column). This system knows the green arrow (because it corrects the AWA for the heel), but it does not know the red arrow, and uses instead the broken arrow. For him, the blue arrow starts at the right position, but ends at the tip of the broken arrow instead of the tip of the red arrow. This is why it will reports a TWA a little less than 45.0 deg, and a TWS a little less than 12.0 knots. By consequence, it will also calculate a different VMG and WDIR.<br /><br />Finally, the first column (System A) is calculated by removing the correction of AWA for heel. For that we need to modify the angle of the green arrow to -26.71 deg (the uncorrected value). The length of the green arrow (AWS) remains the same. So for this system, the blue arrow starts at a different position, and ends again at the tip of the broken arrow. The results are then recalculated for these conditions.Merlinhttps://www.blogger.com/profile/00901116173524809046noreply@blogger.comtag:blogger.com,1999:blog-4237678261594285909.post-54317019912208270432011-02-19T14:06:50.104-05:002011-02-19T14:06:50.104-05:00When calculating the 3rd column, I get a different...When calculating the 3rd column, I get a different value for TWA (52.05) that's about the 45 degrees wind, together with the 7 degrees leeway. VMG = 3.72<br /><br />I used a true wind to AWA excel sheet, I got it somewhere from the internet.<br /><br />Also isn't MBS the same as STW?Erikhttps://www.blogger.com/profile/10896128673707323955noreply@blogger.comtag:blogger.com,1999:blog-4237678261594285909.post-32544393067087458182011-02-19T10:26:30.282-05:002011-02-19T10:26:30.282-05:00Hi Erik,
Heel is 20 deg, K is 12.6
so that atan(ta...Hi Erik,<br />Heel is 20 deg, K is 12.6<br />so that atan(tan(26.71)/cos(20)) = 28.17 deg.<br />and leeway = 12.6 * 20 / 6 / 6 = 7 deg.<br />(Here K is defined by using measured boat speed, not STW).Merlinhttps://www.blogger.com/profile/00901116173524809046noreply@blogger.comtag:blogger.com,1999:blog-4237678261594285909.post-38487586197372020222011-02-19T07:07:43.509-05:002011-02-19T07:07:43.509-05:00Hi Merlin,
Could you post the angle of heel, and ...Hi Merlin,<br /><br />Could you post the angle of heel, and the factor K you used to calculate this?<br /><br />ThanksErikhttps://www.blogger.com/profile/10896128673707323955noreply@blogger.com