Re: boat sways when towing
I found this out on the web and I am still working through the computations to see if I can verify if this is real world usable:<br /><br />Repeat: Not my Work.<br /><br />lets see if I can explain this calculation - this message may be a tad long though. I placed (assumed) the transom at the back of the trailer. Denoting the length of the trailer as X3, the distance from the transom to the center of the wheels as X1, the distance from the center of gravity to the transom as X2, the total weight of the boat/motor/trailer/et.al as Lt, the weight of the boat/motor as Lb and the hitch load as Lh.<br />Now, the bending moment at the transom (back of the trailer) must be zero by definition. Then, writing the summation of moments about the transom (defining clockwise as positive) gives:<br /><br />X3(Lt) + (Lt - Lh)X1 - Lb(X2) = 0<br /><br />and the same basic equation for the modified configuration - same loads but different distances as:<br /><br />Y3(Lt) + (Lt - Lh)Y1 - Lb(Y2) = 0<br /><br />Notice the load on the wheels is written as (Lt- Lh). <br /><br />Now, from the transom, the bending moment of the boat/engine will not change - or namely, Lb(X2) = Lb(Y2)- so strictly speaking, we don't require the location of the center of gravity of the boat/engine. Remember that everything is referenced to the transom.<br /><br />Solving each of the above equations for Lb(X2) or Lb(Y2) - remember, these are the same - and equating those equations gives:<br /><br />LbX2 = X3Lh + X1(Lt - Lh)<br />and<br />LbY2 = Y3Lh + Y1(Lt - Lh)<br /><br />and equating these equations gives:<br /><br />X3Lh + X1(Lt - Lh) = Y3Lh + Y1(Lt - Lh)<br /><br />and solving for Y1 (the distance from the transom to the centerline of the wheels) in the modified configuration gives:<br /><br />Y1 = [X3Lt + (Lt-Lh)X1 - Y3Lh]/ (Lt - Lh)<br /><br />which, when rearranged gives:<br /><br />Y1 = [Lt(X3-Y3) + (Lt-Lh)X1] / (Lt - Lh)<br /><br />and with X3 = 300, Lt = 4700, Lh = 300, X1 = 58, Y3 = 285 i.e. (300-15 : from moving the boat forward 15 inches) gives:<br /><br />Y1 = 59 inches. Which is the distance from the transom to the centerline of the tandem axles in the modified configuration.<br /><br />Note, the original distance from the transom to the centerline of the tandem axles was 58 inches (300 - 242). Therefore, the tandem axle should be placed 59 inches forward of the transom - or in other words, moving the boat forward 15 inches should be accompanied by moving the centerline of the axles 16 inches forward.<br /><br />The above calc considers the transom at the end of the trailer - a few inches one way or the other will not make much difference.<br /><br />Calculating the center of gravity of the boat/engine/et.al can be obtained by solving one of the initial equations, or:<br /><br />from<br />X3Lt + (Lt-Lh)X1 - LbX2 = 0<br /><br />and solve for X2 as:<br /><br />X2 = [X3Lt + (Lt-Lh)X1] / Lb<br /><br />which gives: X2 = 98.63 so the boat/engine center of gravity is 98.63 inches forward of the transom.<br /><br />Everyone should, as desired, check the above calculations. Should you have questions, let me know. I have, rounded some number off - for example, the above 59 inches is actually calculated as 59.024 inches.<br /><br />Now, regarding some of the wags/guesses etc, remember that the boat is the controlling component at 3500 pounds of the total 4700 - so if I were NOT making calculations, I would move the axles the same distance as I wanted to move the boat.