wood specific gravity

  • can we use the specific wood gravity to estimate required balancing leads?
    ex. iroko is 0.64 so w expecting 0.36 weight ratio ,2 kg of wood need 720 g to get neutral =500g spear + SS accessories + lead if needed


    is it near the accurate or far a way ?

  • Not sure I'm understanding you note / question. I don't see the volume of wood, for water displacement value.


    I think most float the gun in water and tape the lead on top in different positions and amounts.


    Not the easiest problem to figure out with a pencil, slide rule and big chief notebook.

  • Not sure I'm understanding you note / question. I don't see the volume of wood, for water displacement value.


    I think most float the gun in water and tape the lead on top in different positions and amounts.


    Not the easiest problem to figure out with a pencil, slide rule and big chief notebook.


    Good point. Make sure you have the math correct before proceeding.

  • The math is a bit off since, as Linghunt points out, Mosaad is not taking volume into account.


    But Mosaad does actually have an estimated volume available to him by way of the weight of the shaped blank (2000g) and the specific gravity of 0.64g/cm3. I guess some woods' gravity can range a lot and I don't know if the 0.64g/cm3 is one Mosaad calculated himself from measuring his wood or took from a table of densities.


    Anyways, if we assume that the 2 kg is the weight of the shaped blank and the 0.64g/cm3 is the correct density of the wood he is using, then the volume of his blank is (2000g / 0.64g/cm3) = 3125cm3.


    The handle will add volume and the shaft and rubbers, too so the overall volume will be bigger.
    The volume of two 16mm bands for a gun with 110 cm band stretch is about 270cm3 (using 340% stretch). Perhaps a handle is about 150cm3? (no idea, really). A 140cm, 7mm shaft would be around 54cm3.
    If we use these figures, the total volume of the gun would be (3125+270+150+54cm3) = 3599cm3.


    So, in freshwater, the whole gun should weigh about 3.6kg to be neutral.
    But I found an old number for the density of sea water in Abu Dhabi and it said it was 1.029g/cm3. So, I guess, in theory, Mosaad's gun should be neutral in Abu Dhabi waters if it weighs around (3599cm3 x 1.029g/cm3) = 3703g. Let's call it 3.7kg;-).
    No matter what, it sounds to me like a beefy gun for a euro?


    Yeah, I had nothing better to do this morning:-)
    Also, as always, my math may not be correct, haha.


    [EDIT]
    Obviously, I took the liberty of guessing he is making a two-banded 110cm gun but I can see now that he quotes his spear weight at 500g, which points to a larger gun. Anyways, I think the method in itself is sound and the volume numbers would not change much either way (just different bands and spear volumes).
    But you can also just build the gun and ballast/balance it in the pool, like everyone else;-).

  • True, if you CAD your design with all the parts you can pretty easily get your total volume and then multiply that with the density of salt water and you will have a mass that your gun should have if you want it neutrally ballasted.
    Now, balance is another thing and a tad harder to find in CAD.
    You can still get the center of volume easily but to get the center of gravity you need to specify the density of all the different parts in your gun; wood, steel, rubber, etc.


    Ideally, you would want the center of gravity aligned with the center of volume or buoyancy - then the gun would float level. You can also make it slightly nose heavy, some folks like that.


    But remember, that while you can add your lead ballast in one spot to place the center of gravity where you want it, you can achieve the same center of gravity by dividing the lead with some in the front and some in the back. It seems most builders prefer this. Not only because it gives them more room to dial in the balance (nose up or down) but also because if you place weight at the extreme ends, the gun will have more inertia and should, in theory, resist muzzle flip more.

  • if you want to make your stock ready with balanced drilled holes before going to the sea to get the accurate lead needed I just make very simple experiment and want to share it to everyone here ,


    need simple tools :big water jar , your wood sample ,weight and digital scale


    attach the weight to your wood till sink or what ever then measure the required weight and multiply to your gun weight
    ex. I have padouk and billinga
    padouk weight is 69 gram will sink with 25 gram of ss ratio is 0.3623
    billinga weight 96 gram and need 41 gram ss ratio is 0.427
    so when I make gun with 1kg padouk I need approxi. or around 362 gram
    gun with 1.5kg billinga I need approxi. or around 640 gram


    Is that correct and how to calculate for seawater ?

  • I guess you could do it like that. Just remember that you don't need all the weight in lead. Your spear and trigger and other metal parts should be subtracted from the lead mass.


    For sea water, you should multiply your total mass of the neutral assembly (wood, lead, metal parts) by approximately 0.025-0.029 depending on salinity of your area and that should give your the amount of lead you need to add.


    But why don't you do something else - and what I mentioned earlier?
    Measure the weight and volume of your wood samples. That will give you the exact specific gravity of the wood you are using.
    Then you measure the weight of your final shaped wooden gun (without metal parts) and divide it by the specific gravity and that will give you the volume of your gun. Let's say your volume is 1500cm3. Then add the volume of the spear. E.g. a 140cm, 7mm shaft would be around 54cm3.
    You don't really have to take into account the rubbers as they are almost neutral (very slightly buoyant).
    So, the volume of you gun and spear is (1500+54): 1554cm3.
    In fresh water, the gun should weigh the same in grams to be neutral. So, your gun with a volume of 1554cm3 should weigh 1554 grams to be neutral.
    For sea water, multiply by 1.025-1.029 and that will give you the total mass to make the gun neutral.


    Eg. your 1554cm3 gun should weigh (1554cm3 x 1.025g/cm3): 1593g to be neutral in seawater (with a water density of 1.025g/cm3). If your water is very salty you might need more weight.


    Finally, assemble your gun with all the parts and weigh it - add lead until you get to 1593g. Done.


    (If you are mixing wood in your blanks, it might not be as easy as the above)

  • Every piece of wood is different: That is why specific gravity is generally listed as a range rather than an exact value. Math can give you a good approximation, but perfect weight and, equally important, balance, are best achieved in the water

  • Every piece of wood is different: That is why specific gravity is generally listed as a range rather than an exact value. Math can give you a good approximation, but perfect weight and, equally important, balance, are best achieved in the water


    Which is why I proposed he measure a sample of the wood that he is using and not just rely on a list. With a ruler and a scale you can easily find pretty much the exact density of the wood you are using to make your gun out of. This should get you some quite usable results if you are far from a pool or ocean.

  • Every piece of wood is different: That is why specific gravity is generally listed as a range rather than an exact value. Math can give you a good approximation, but perfect weight and, equally important, balance, are best achieved in the water


    True, but good for a base point. To get an idea at least.

  • Holy hell Diving gecko, you are a math genius. Thank you for that. Is it tried and true?


    Only tried in my mind...:@:D
    So, I am really hoping that some of the engineers or math guys here will double check it. Pete could see through any issues with the method in a split sec so let's hope he chimes in:-). That said, I am sure enough of the math that I would use it myself for what it is worth;-)


    I am landlocked, but without a workshop and have a quarter of an engineer stuck inside my photographer's body longing to get out - furthermore, I have too much time on my hands sometimes so I tend to have to solve spearing puzzles in my mind way more than in real life, haha.

  • I used the terrestrial method on my last two bluewater guns. Came out fine with no need for further adjustments. Just make sure your math and methods are good.:thumbsup2::toast1::thumbsup2:

  • I would not get too hung up on what method you use to ballast. Probably better to do it in the water the first few times. If you design your gun to allow for easy adjustments to ballast it could be a benefit if you change shaft size or terminal gear. I am really not all that picky as long as its in the ball park. I am usually so stoked just to be in the water that an ounce here or there is not a factor.

  • My point was that math can get you close, but knowing roughly the specific gravity of the blank, you can build the gun you want, and adjust buoyancy to fit the exact properties of your gun.


    And my point is that if you are using only one type you don't need to settle for "roughly":). You can measure the size and weight of a piece of the wood you are using and obtain the exact density of that wood. It would take about a minute. Unless the density changes within your batch, then you will have very good numbers to work with.
    So, in terms of buoyancy you should be able to get it pretty much perfectly neutral without even having access to water.


    Now, fore and aft balance let think about that one. Perhaps, it is a simple as balancing the rigged up gun on a "knife's edge" and moving the lead around until the gun is level. The gun as a whole, with the ballast, should have the same density as water so this might just work. Now, if you want to dial in any nose heaviness, I guess that's for the pool. Overall, I still think you can get extremely close in your workshop:)

  • I d like to update one thing ,after experiment I found out after using epoxy to glue the padouk the floating ratio I calculated increase ! I mean the required weight to sink the specimen is different when u laminate the piece padouk ratio increased to 0.41 ,the epoxy glue have air bubbles which impossible to eliminate it 100% so is that the main reason? uncountable air bubbles trap air inside increase the flotation ?

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