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Vlanikgun

tromic

Well-Known Member
Aug 13, 2007
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To increase reliability of "sealed" tail cap design I would make something like this on drawing. Sealing o-ring is shielded from scratches as seen on the picture. Rubber plug makes friction to the cap so ensures more reliability in case of o-ring sealing failure. Air bubbles after shooting could indicate bad sealing.

This would be a mixture with Vladimir's design. I believe design with neoprene would also work.

There is an another problematic statement regarding performance of Vlanikgun. I do not believe that 28 cm long gun can have 2,8 m effective range. That range is obtained by calculation taking initial speed of shaft in air to be same in water, what is not true. According to that calculation, 70 cm long gun has range 5,6 m. That gun has about the same shaft acceleration distance as my Cyrano 850 (64 cm). My Cyrano (Tomba) with 100/7 mm shaft has effective distance to 4 m. Vlanik 70 might have to 4,5 m because of thicker shaft, not 5,6 m.
 

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Vlanik

Well-Known Member
Jan 19, 2009
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Не нужно верить, нужно проверять...
=============================
Not it is necessary to believe, it is necessary to check...
 

tromic

Well-Known Member
Aug 13, 2007
1,672
206
153
Не нужно верить, нужно проверять...
=============================
Not it is necessary to believe, it is necessary to check...

Yes, you are right, I agree Vlanik, would be nice if I could check it. But 2,8 m effective range for 28 cm long gun is maybe too much, on 25 kg loading effort? I do not think effective range might be more than 2 m. 2 m would be very good too!
 

Vlanik

Well-Known Member
Jan 19, 2009
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Ружья способны стреллять на дистанции равной 10 длин ружья...(5 витков линя)
Оптимально применяемая дистанция 8 длин ружья...(4 витка линя)
Нельзя делать линь короче 6 длин ружья...(3 витка линя)
Если делать (2 витка линя) гарпун возвращается и попадает в стрелка...
=============================================
The Handguns capable стреллять on distances equal 10 lengths of the handgun...(5 whorls линя) Optimum applicable distance 8 lengths of the handgun...(4 whorls линя) It is impossible do линь shorter 6 lengths of the handgun...(3 whorls линя) If do (2 whorls линя) harpoon returns and falls into arrow...
 

tromic

Well-Known Member
Aug 13, 2007
1,672
206
153
Ружья способны стреллять на дистанции равной 10 длин ружья...(5 витков линя)
Оптимально применяемая дистанция 8 длин ружья...(4 витка линя)
Нельзя делать линь короче 6 длин ружья...(3 витка линя)
Если делать (2 витка линя) гарпун возвращается и попадает в стрелка...
=============================================
The Handguns capable стреллять on distances equal 10 lengths of the handgun...(5 whorls линя) Optimum applicable distance 8 lengths of the handgun...(4 whorls линя) It is impossible do линь shorter 6 lengths of the handgun...(3 whorls линя) If do (2 whorls линя) harpoon returns and falls into arrow...

28 cm x 8 = 224 cm. I was close to this with my estimation of 2 m.
How much energy (J), have you on optimal distance? Can the shaft go through the fish of 3 - 4 kg on optimum distance?
 

Vlanik

Well-Known Member
Jan 19, 2009
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Практически все стрелляют с более коротких дистанций чем позволяет линь...
Поэтому гарпун проходит навылет...
Все характеристики для ружей можно расчитать в EXSEL...На моём сайте...
Что касаемо веса рыб и требуемой для их поражения энергии, у меня нет данных...
Никто не хочет этим заниматься...
Хотя это легко определяется...
==================
Practically all стрелляют with more short distance than allows линь...
So harpoon passes going right through...
All features for ружей possible расчитать in EXSEL...On my put...
That касаемо weight of fish and required for their defeats of the energy, I have no data...
Nobody does not want this concern with...
Though this is easy defined...
 

tromic

Well-Known Member
Aug 13, 2007
1,672
206
153
Практически все стрелляют с более коротких дистанций чем позволяет линь...
Поэтому гарпун проходит навылет...
Все характеристики для ружей можно расчитать в EXSEL...На моём сайте...
Что касаемо веса рыб и требуемой для их поражения энергии, у меня нет данных...
Никто не хочет этим заниматься...
Хотя это легко определяется...
==================
Practically all стрелляют with more short distance than allows линь...
So harpoon passes going right through...
All features for ружей possible расчитать in EXSEL...On my put...
That касаемо weight of fish and required for their defeats of the energy, I have no data...
Nobody does not want this concern with...
Though this is easy defined...

From excel data on your site I read that on optimum distance for your gun you have shaft momentum (impuls harpuna) about 0,8 kg*m/s. Energy for two guns on optimum distance are very different. So momentum is more important than kinetic energy to determine the effective shooting range? But, 0,8 kg*m/s to me seems to be not enough? I suppose there should be about 5 J of kinetic energy on target and 10 - 15 m/s shaft speed. Momentum should be at least 1,2 kg*m/s for short gun or more for longer gun, at least 3,5 kg*m/s for 90 cm gun. Peter, what do you think?
 
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Vlanik

Well-Known Member
Jan 19, 2009
150
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Для поражения рыбы весом 1 кг., достаточно импульса 0,6 кг*м/с...
Пользуясь exsel, следует учитывать, что внешняя баллистика считается с того момента, когда гарпун покинул ружьё...А это значит, что дистанция поражения увеличивается на длинну разгонного (рабочего) хода гарпуна...
=======================
For defeat of fish by weight 1 kg., it is enough pulse 0,6 kg*m/with...
Using exsel, follows to take into account that external ballistics is considered from that moment, when harpoon has abandoned the handgun...But this signifies that distance of the defeat increases on long runaway (the worker) of the move of the harpoon...
 

tromic

Well-Known Member
Aug 13, 2007
1,672
206
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If I have a shaft of 0,1 kg than for momentum 0,6 kg*m/s, speed must be 6 m/s.
It is same as if I let a shaft with mass 0,1 kg to fall down to the fish from 1,8 m height. The speed will be 6 m/s. Right? I think that is not enough to get through the fish of 1 kg.
 

Vlanik

Well-Known Member
Jan 19, 2009
150
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Vlanik-20


The most small undersea gun!

The Idea of the creation of such small gun was born in fun
Someone have said: - Fish with laughter will surface up брюхом, when his(its) will see I have offered Dmitriyu to try that is thereof got
And here is his(its) first impressions
=============================
Either as promised, report . In spite of april fools day establish in not карповых places Vlanik-20 is threatened become main handgun for me.
The Ithaca
The River of the Court, air +3, water +1.5 бревна Zanyrnul, has done the full shot. Usual for влаников CIK and harpoon has left for limit of visibility by thread. Lini -2 metres, on hand has felt the blow . The Metre on 2.5 will exactly fly.
Sails . The Truth on mark -0.5 metre . The Lung, in log and завалах ideal size .
Here and налимчик was drawn - has directed, has pressed was not pressed Turned out to be the fuze a little upwards rose. Has Put(deliver)ed on place and стрельнул .. on кукан him(it)
подлещик shot with metre полутора .. has not got.
the окуньков small flock while выцеливал large разошлись fell into 100 gr. Otplaval vapour(pair) hours, has looked at tails vapour(pair) descending pike perch and лег on inverse course.
Passing the log see in полводы under itself . The Pike perch! Turned round, вздохнул and зашел overhand with tail hand has extended and has snapped . Feel got! Adrenaline in shelters, review on pike perch something unalike .. pike!
Has Overpunched overhand her spine, with metre sometime, not through, outward protruded only roller. Dotolknul harpoon and on кукан her(it) . Was necessary see the sad eye my дружков, вылезших empty.
Here is she beauty.
3,8 kgs 80 refer to
Hiccupped: 2 hours, four shots, three fish .
 
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popgun pete

Well-Known Member
Jul 30, 2008
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From excel data on your site I read that on optimum distance for your gun you have shaft momentum (impuls harpuna) about 0,8 kg*m/s. Energy for two guns on optimum distance are very different. So momentum is more important than kinetic energy to determine the effective shooting range? But, 0,8 kg*m/s to me seems to be not enough? I suppose there should be about 5 J of kinetic energy on target and 10 - 15 m/s shaft speed. Momentum should be at least 1,2 kg*m/s for short gun or more for longer gun, at least 3,5 kg*m/s for 90 cm gun. Peter, what do you think?

I do not see calculations as anything but an approximation, particularly as shooting line texture and line thickness influences line drag and affects the speed and flight distance of the shaft. If you shoot "sitters", fish lying on the bottom in ambush, then the shaft speed does not have to be very high as it strikes the fish provided that the shaft is heavy and the gun is quiet. Shooting downwards helps if the situation allows it. Fish on the move need a faster spear to intercept their course, plus are not being impaled while being jammed up against the bottom. So effectiveness of gun can depend on the hunting circumstances. I note that shooting line diameter indicated on spreadsheet is 1.00 mm, very small!

The energy stored in the gun is the most important factor, followed by the efficiency in how much of that energy is converted to spear velocity. The "Vlanik" gun data shows that efficiency is 0.94 or 94%, so if the gun performs better then it must be due to the higher efficiency. Calculation of distance needs a number of factors to be estimated and may be different from the reality depending on the validity of the assumptions used in the calculations, however ranking of guns may be correct for the same operating conditions. Freshwater versus saltwater may have an effect due to different density, but I have never calculated it. Not sure what КПИ indicates, the value is less than 1.0 for each gun on the datasheet.
 

tromic

Well-Known Member
Aug 13, 2007
1,672
206
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Vladimir, can you write me a formula how do you calculate the energy of the shaft for different distances in your excel table? You have in your table (blue):
Distance____Energy
0,250______21,825
0,500______15,060
0,750______10,392
....
 

Vlanik

Well-Known Member
Jan 19, 2009
150
25
118
E=(M*V*V)/2=(0,11762*19,264*19,264)/2=21,8249......J
Программа делает подобный расчёт через каждые 50 мм. дистанции полёта гарпуна...
В таблице этот расчёт показан через каждые 250 мм. полёта гарпуна...
===============================
The Program does such a calculation through each 50 mm. distances flight harpoon...
In table this calculation is shown through each 250 mm. flight of the harpoon...
 

tromic

Well-Known Member
Aug 13, 2007
1,672
206
153
E=(M*V*V)/2=(0,11762*19,264*19,264)/2=21,8249......J
Программа делает подобный расчёт через каждые 50 мм. дистанции полёта гарпуна...
В таблице этот расчёт показан через каждые 250 мм. полёта гарпуна...
===============================
The Program does such a calculation through each 50 mm. distances flight harpoon...
In table this calculation is shown through each 250 mm. flight of the harpoon...

OK, so you calculate energy from speed. Than better question would be how you calculate the speed (v)?
 

Vlanik

Well-Known Member
Jan 19, 2009
150
25
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OK, so you calculate energy from speed. Than better question would be how you calculate the speed (v)?
Просто всё делается...
Определяется разгоняющая сила...
Определяются силы сопротивления...
Определяется путь разгона...
Определяется скорость на участке разгона...
В алгоритме определения скорости, таких участков разгона 100...
Это даёт точность вычисления +- 1 процент...
Вот алголритм расчёта скорости на 1/10 части общего пути разгона с учётом КПД...
=================================================
Simply all are done...
It Is Defined dispersing power...
Power of the resistance are Defined...
The way of the runaway is Defined...
The velocity is Defined on area of the runaway...
In algorithm of the determination to velocities, such area of the runaway 100...
This gives accuracy of the calculation +- 1 percent...
Here is алголритм calculation to velocities on 1/10 parts of the general way of the runaway with account KPD...

 
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tromic

Well-Known Member
Aug 13, 2007
1,672
206
153
Просто всё делается...
Определяется разгоняющая сила...
Определяются силы сопротивления...
Определяется путь разгона...
Определяется скорость на участке разгона...
В алгоритме определения скорости, таких участков разгона 100...
Это даёт точность вычисления +- 1 процент...
Вот алголритм расчёта скорости на 1/10 части общего пути разгона с учётом КПД...
=================================================
Simply all are done...
It Is Defined dispersing power...
Power of the resistance are Defined...
The way of the runaway is Defined...
The velocity is Defined on area of the runaway...
In algorithm of the determination to velocities, such area of the runaway 100...
This gives accuracy of the calculation +- 1 percent...
Here is алголритм calculation to velocities on 1/10 parts of the general way of the runaway with account KPD...

Not helped, but thanks. Maybe Peter understand this table.
 

popgun pete

Well-Known Member
Jul 30, 2008
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Not helped, but thanks. Maybe Peter understand this table.

I had already attempted to re-create Vladimir's first XL spreadsheet, as you already know there are no formulas in the downloaded example from his web-site. Once the spear leaves the muzzle then the only forces acting on the shaft are the resistive force of the water (horizontal) and gravity (vertical) on the shaft itself and the drag from the shooting line trailing along behind it. At moment of separation of shaft tang from tail cap there is also a brief frictional resistance. I assume that the horizontally acting factors are somehow incorporated in the "Opposing Force Fc" column, i.e. third column from the left, not including the empty column. Shaft energy and impulse/momentum calculation is dependent on the instantaneous velocity at each distance interval from the muzzle, however instantaneous velocity must first be calculated from the decelerating effect of the "Opposing Force".

Separate "Calculation of velocity V" table does not show the algorithm used, I note that calculation is to 6 figures for "Dispersal Force Fp" and "Opposing Force "Fc"! Insufficient information as to the algorithm used, but it must have a number of input variables not shown here. Velocity at each distance interval is sum of velocity at previous interval plus value of velocity increment dV (delta V) for that interval or V = V + dV. Values of V and dV are somehow calculated from "Dispersal Force Fp" and "Opposing Force Fc", last two columns respectively on "Calculation of velocity V" table. Values of X are incremented in steps of 0.001, so this "Calculation of velocity V" table must be very large! More information is required if you want to know how the calculations are made.
 
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popgun pete

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Jul 30, 2008
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Starting point for any calculation is the energy stored in the gun after pushing the spear into the gun. This energy is determined by the initial chamber pressure before loading and the final chamber pressure after loading, the working stroke of the spear and the cross-sectional area exposed to the varying chamber pressure as the shaft is pushed into the gun. Initial pressure and final pressure are governed by the gun's compression ratio. The energy stored in the gun is equivalent to the area under a graph of chamber air pressure varying with shaft travel inside the gun multiplied by the cross-sectional area of the spear. Energy stored is derived from the changing force level (chamber pressure times shaft cross-sectional area) applied over the distance travelled by the object being moved which is the spear. The area under the graph is the equivalent of a rectangle with a right angle triangle sitting on top of it, base of the triangle is the same length as the rectangle, in this case 15 cm for small gun (working course of the spear which is same length as reservoir according to table).

I calculate the compression ratio as 1.2154 (refer my earlier post). At maximum chamber pressure 49.74 atm (for already loaded gun) the maximum force to latch is 25 kg, so initial chamber pressure in gun will be given by 49.74/1.2154 (final chamber pressure divided by compression ratio) which equals 40.92 atm. Force to get spear started (neglecting friction both at muzzle seal and insertion of shaft tang in tail cap) will be 40.92 atm multiplied by shaft cross-sectional area (0.503 sq cm) which equals 20.57 kg (assuming here atm is same as kg/sq cm). For shaft travel distance of 15 cm the stored energy then calculates as 34.87 which compares with 31.62 (from table) at muzzle in terms of shaft kinetic energy, so we have lost 3.25 Joules (unit of energy) or about 9% of the stored energy. Surprising result as I was expecting only 6% loss given the efficiency figure of 94%. Stored energy result may be overstated if force to load was partly consumed in overcoming the friction that I have neglected in the calculation, in which case the efficiency will improve. After the spear leaves the muzzle the energy it retains during shaft flight determines how far it flies and the velocity, the gun then becomes irrelevant as its job has been done.
 
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popgun pete

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Jul 30, 2008
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I found the 2005 diagram in my old computer, I had drawn it up using the simple drafting tools in MS Word, but it is now converted to an image file here. The diagram is only a schematic, not an engineering drawing, so the rear "power dial" would be more complicated than it is shown here. Like the "Vlanik" gun at the time the gun is pressurized through the muzzle with a wand type pump fitting locked into the muzzle, there is no air inlet valve at the rear end of the gun.
 

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