Stability of Paddle Boards at the Lake
William George
Marine Educator, Consultant and Expert Witness at William E. George and Associates, LLC
Standup Paddle Boards have become very popular. While on vacation this summer I started to watch people on Paddle Boards and decided in my lawn chair at the lake to investigate their stability. Why? Just because I like numbers and I can so why not?
Using the same equations that I use for ships in my book STABILITY AND TRIM FOR THE SHIP’S OFFICER, 3 and 4th Editions, it is easy to study smaller craft by using appropriate units such as Pounds instead of Long Tons and Inches instead of Feet. I know I should have done this in Metric Units, but I am an American and it is Summer, and I am on Vacation. In no way I work or endorse any Paddle Board products. I thought why not see what the numbers tell me. I would like to stress the following addresses Transverse Stability only. I realize that Paddle Boards are Longer and Narrower it will be much easier to propel and maneuver but that is not the point of the following.
?To keep things simple, let us explore the Stability of a Paddle Board with a 6-foot tall 150-pound person standing on a rectangular Paddle Board measuring 8 feet long, 3 feet wide, a depth of 0.5 feet and weighs 31.2 pounds. Note: most Stand Up Paddle boards are 9 to 11 feet long with tapered ends, I chose an effective length of 8 feet for this evaluation of transverse stability only.
?First let us calculate the draft with the person stand at the LCB, Longitudinal Center of Buoyancy. This should be a zero-trim condition, so parallel sinkage rules.
?Based on the overall dimensions the total weight the Paddle Board could support and still float is equal to:
?8 ft x 3 ft x 0.5 ft = 12 cubic feet displaced volume. Allowing 62.4 pounds per cubic foot of fresh water the maximum weight is equal to: 12 cubic feet x 62.4 lb/ cu ft =?748.8 pounds.
The “Pounds per Inch Immersion” will be equal to: 748.8 lb / 6 inched = 124.8 lb / in.
?If the empty Paddle Board weighs 31.2 pounds the light draft will equal 0.25 inch. When the 150-pound person boards the paddle board the increase in draft will be equal to:
150 lb / 124.8 lb/in =?1.2 inches
?So, the loaded draft will be equal to:
0.25 in + 1.2 in = 1.45 inches
?The Loaded Displaced Weight will be equal to draft x pounds/ inch: Paddle Board + Person = 31.2 + 150 lb/in = 181.2?pounds.
?For Fresh Water the Displaced Volume = 181.2 lb / (62.4 lb / cu ft) = 2.903 cubic feet. We will need this later.
?Initial Stability can easily be measured by using GM.
?GM = KM - KG
KM = KB + BM?
If GM is "positive' the Paddle Board should return to its original condition when disturbed by an outside force.
If GM is "negative" is will not float upright and have an "Angle of Loll" or even capsize!
First, we need to calculate KB, the "Center of Buoyancy" where K is the lowest point on the Paddle Board. Because this Paddle Board has a rectangular shape KB will equal 1/2 the Loaded Draft which will be equal to:
KB = 1.45 inches / 2 = 0.725 inches or 0.06 feet.
?Now we need BM, the "Metacentric Radius"
?BM = I, "Moment of Inertia of the Transverse Water Plane" divided by V, the "Displaced Volume"
?BM = I / V
?I = length x (beam x beam x beam) / 12
?I = 8 ft x 3 ft x 3ft x 3ft / 12?= 18 ft to the 4th power
?V was previously calculated for the loaded condition = 2.903cu ft
So, BM = 18 ft ^4 / 2.903 ft^3 = 6.20 ft
(Note: "feet to the 4th" divided by "feet cubed or to the 3rd" = feet.)
KM = KB + BM?
KM = 0.O6 ft + 6.20 ft = 6.26 ft
?GM = KM - KG
So now we need to calculate or estimate "KG" of a 6-foot-tall person standing on the Paddle Board. Allowing for the trunk of the human body, arms and head weighs more than the legs let us assume the center of gravity is 2/3rds of the height or 4 feet up from the feet. To be as accurate as we can, we must calculate a "weighted average" including the weight and height of the center of gravity of the Paddle Board itself and the weight and the center of gravity of the height of the human.
?The vertical moment of the board is equal to:
?31.2 pounds x O.25 ft = 7.8 ft-lb
The vertical moment of the human standing is equal to:
?150 pounds x (4.O ft + 0.5 ft) = 675 ft-lb
?Summing the vertical moments, we get:?
?7.8 ft-lb + 675 ft-lb =?682.8 ft-lb
?Summing the weights, we get:
?31.2 lb + 150 lb = 181.2 pounds
?KG = Vertical Moments / Weight
KG = 682.8ft-lb / 181.2 lb = 3.74 ft above "K", the keel or the plane of the baseline of the Paddle Board.
GM = KM – KG = 6.26 ft - 3.74 ft = + 2.52 ft.
Conclusion: Paddle Board is Initially Stable for only Small Angles of Heel.
For this condition what is a safe Angle of Heel?
Maxim Righting Arm occurs a very little past "Deck Edge Immersion". Let's keep it simple and say it is at "Deck Edge Immersion".?I propose 1/2 the Freeboard would be a reasonable safe limit. So, what is the Freeboard?
?Freeboard = Depth - Draft?
The Depth is 0.5 feet or 6 inched and the Loaded Draft is 1.45 inches.
?Freeboard = FB = 6 in - 1.45 in = 4.55 inches
1/2 Beam = 3 feet/2 or 36/2 inches = 18 inched.
Tangent (Safe Angle of Heel) = ? FB / (Bean/2) = 2.275 / 18 = 0.12638
Safe Angle of Heel = arctan (0.12638) =?7.20 degrees.
?Judgement. Relatively large Safe Angle of Heel. The Paddle Board is not overloaded with a reserve buoyancy of 567.6 pounds.
?Now, let us Consider the Stability of the Paddle Board with a 300 pound, six-foot-tall human or two six-foot-tall 150-pound-humans standing on the Paddle with everything else the same.
?As before, this should be a zero-trim condition, so parallel sinkage rules.
?Based on the overall dimensions the total weight the Paddle Board could support and remain afloat is unchanged.
8 ft x 3 ft x O.5 ft = 12 cubic feet displaced volume. Allowing 62.4 pounds per cubic foot of fresh water the maximum weight is equal to: 12 cubic feet x 62.4 lb/ cu ft =?748.8 pounds. The Pounds per inch immersion will be equal to: 748.8 lb / 6 inched = 124.8 lb/in.
?If the empty Paddle Board weighs 31.2 pounds the light draft will equal 0.25 inch. When the 300-pound-person boards the paddle board the increase in draft will be equal to:
300 lb / 124.8 lb/in =?2.4 inches
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?So, the loaded draft will be equal to: Parallel Sinkage of the Paddle Board and the Person = 0.25 in + 2.4 in = 2.65 inches
?The Loaded Displaced Weight will be equal to draft x pounds/ inch: Paddle Board + Person = 31.2 lbs + 300 lbs = 331.2 pounds.
?For Fresh Water the Displaced Volume = 331.2 lb / (62.4 lb / cu ft) = 5.31 cubic feet. Again we will need this later.
?Initial Stability can easily be measured by using GM as previously discussed.
?GM = KM - KG
KM = KB + BM?
If GM is "positive' the Paddle Board should return to its original condition when disturbed by an outside force.
If GM is "negative" is will not float upright and have an "Angle of Loll" or even capsize!
?The "Center of Buoyancy" is measured from K, the lowest point on the Paddle Board upward. Because this Paddle Board has a rectangular shape KB will equal 1/2 the Loaded Draft which will be equal to:
KB = 2.65 inches / 2 = 1.325 inches or?0.11 feet.
?Now we need BM, the "Metacentric Radius"
?BM = I, "Moment of Inertia of the Transverse Water Plane" divided by?V, the "Displaced Volume"
?BM = I / V
?I = length x beam x beam x beam / 12 = LB^3/12
I = 8 ft x 3 ft x 3ft x 3ft / 12?= 18 ft to the 4th power
?V, displaced volume, was previously calculated for the loaded condition = 5.31 cu ft.
?So, BM = 18 ft^4 / 5.31 ft^3 =?3.39 ft
(Note: "feet to the 4th" divided by "feet cubed or to the 3rd" = feet.)
KM = KB + BM?
KM = 0.11 ft + 3.39 ft = 3.50 ft
GM = KM - KG
So now we need to calculate or estimate "KG" of a 6-foot-tall person standing on the Paddle Board. Allowing for the trunk of the human body, arms and head weighs more than the legs let us assume the center of gravity is 2/3rds of the height or 4 feet up from the feet. To be as accurate as we can calculate a "weighted average" including the weight and height of the center of gravity of the Paddle Board itself and the weight and the center of gravity of the height of the human.
?The vertical moment of the board is equal to: 31.2 pounds x O.25 ft = 7.8 ft-lb
?The vertical moment of the human standing is equal to: 300 pounds x (4.O ft + 0.5 ft) = 1350 ft-lb.
Summing the vertical moments, we get:?7.8 ft-lb + 1350 ft-lb =?1357.8 ft-lb
?Summing the weights, we get: 31.2 lb + 300 lb = 331.2 pounds
KG = Vertical Moments / Weight
KG = 1357.8 ft-lb / 331.2 lb =?4.10 ft above "K".
?GM = 3.5 ft - 4.10 ft = - O.6 ft.
?Judgement: Paddle Board is Initially Unstable if the 6 foot tall 300-pound-human is standing! It will not remain upright if disturbed by an outside force and could easily capsize by a large boat wake. The Paddle Board is not overloaded and has a reserve buoyancy of 417.6 pounds.
If the Paddle Boarder kneels their Center of Gravity will be lowered to 2.5 feet above K. KM will remain the same equal to 3.50 feet because the draft has not changed.
?The sitting GM = KM – KG = 3.50 ft - 2.50 ft = + 1.0 foot.
?Judgement: The 300-pound-person kneeling on the Paddle Board is Initially Stable with a Freeboard of 2.35 inches.
?For this sitting condition what is a safe Angle of Heel?
?Maxim Righting Arm occurs a very little past "Deck Edge Immersion". Let's keep it simple and say it is at "Deck Edge Immersion".?I propose 1/2 the Freeboard would be a reasonable safe limit. So, what is the Freeboard?
Freeboard = Depth - Draft?
The Depth is 0.5 feet or 6 inched and the Loaded Draft is 2.65 inches.
?FB = 6 in - 2.65 in = 2.35 inches, so ? the FB = 2.65 inch / 2 = 1.325 inches
1/2 Beam = 3 feet/2 or 36/2 inches = 18 inched.
?Tangent (Safe Angle of Heel) = FB / (Bean/ 2) = 1.325 / 18 =?0.07361
?Safe Angle of Heel = arctan (0.07361) =?4.36 degrees.
?Judgement. In Kneeling Position the Paddle Board has a relatively smaller Safe Angle of Heel of 4.36 degrees. The Paddle Board is not overloaded and has a reserve buoyancy of 417.6 pounds.
*****
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Bunker and Draft Survey Course
Did you know that a Draft Survey properly done can yield the Total Displacement within 0.5% of the true cargo figure or Displacement? You can determine your vessel's total cargo weight by reading the drafts instead of the cargo manifest which may or may not be correct. Draft Surveys works on all types of vessels as long as they are afloat and can confirm the Stability Calculation.
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Paddle Canada (SUP) Instructor 2015 - 2019 – standuppaddleboardingguide.com
1 年Wow! I haven't thought of stability of a paddle board in this way. Fascinating! I imagine board manufacturers and designers would certainly take all this into account. For those like me reading this and as a consumer of boards, most paddle boarders think about stability from an intuitive perspective: Experienced paddlers and non-expert frequent paddlers will look a board, take into account the shape, note the dimensions and have a very good idea of the stability factor on the first ride. The degree of balance a person has, the skill level, the strength, the knowledge and the ability to control a board also influence how a paddle boarder would interpret the level of stability. Obviously my writing is not an apples to apples comparison (science to intuition). Just wanted say this for all the non-math people out there who are interested in paddle boarding. Thanks!
Ocean Engineer/Naval Architect - Diver - Program Manager
1 年I’ve used similar examples in my salvage class, but have injected a block coefficient, water plane coefficient, and transverse/longitudinal water plane inertia coefficients.
Surveyor and Workboat Consultant.MCA small commercial vessel Coding Surveyor IIMS/RYA/BCTQ. ISM Lead Auditor.
2 年Love this.
President and Principal Advisor at Noble Associates Inc., Offshore, Maritime and Arctic Technology Advisors
3 年I think you have to take another thing into consideration. The paddle boarder (like a kayaker) is not fixed and can and will move her center of gravity as the craft heels. This means that the active load of the paddle boarder may act at least in part like an "reverse" suspended load from a stability point of view, where the line of action of her weight acts through her ankles/
Ocean Engineer/Naval Architect - Diver - Program Manager
3 年As my daughter demonstrated with a recent TikTok challenge - CG (generally speaking) for males is around the sternum - so 2/3*h is a good estimate. For females, the general case CG is around the pelvis which is close to 1/2*h. Applying the same math as before, you can demonstrate that the average female is more stable on a paddle board than the average male of same weight/height.