boat

I'm planning on building a cajun sneak boat. I want to use it for a fishing
situation in a small shallow bay. What is the optimal colour to paint the
submerged portion so it doesn't scare the fish ? It's a nice starter project.

"Gone Angling" wrote in message
...
I'm planning on building a cajun sneak boat. I want to use it for a
fishing
situation in a small shallow bay. What is the optimal colour to paint the
submerged portion so it doesn't scare the fish ? It's a nice starter
project.



How about taking a swim and looking up from under water.


--
Bob La Londe
Yuma, Az
http://www.YumaBassMan.com
Promote Your Fishing, Boating, or Guide Site for Free
Simply add it to our index page.
No reciprocal link required. (Requested, but not required)

Paint it a bright color like Blaze Orange or Hot Pink. The portion you are
referring to is typically called a "hull".
This isn't a joke either, rumor has it bright colored boat bottoms and side
help keep bass from jumping. Which if your in a tournament is a good thing.
I was told that Pro's enjoyed fishing out of the Kellogs boat because the
bass didn't jump as much.

Regardless of what color you paint it on somedays it will stand out on some
days as weather, water clarity, and depth of the fish all have an impact on
what the fish see's. An old timer who owned a lot of boats told me that he
caught more fish out of a red boat than any other color. He fished the same
couple of lakes for years and used that as his gauge, luckily I like red

---
Chuck Coger
http://www.fishin-pro.com


"Gone Angling" wrote in message
...
I'm planning on building a cajun sneak boat. I want to use it for a
fishing
situation in a small shallow bay. What is the optimal colour to paint the
submerged portion so it doesn't scare the fish ? It's a nice starter
project.

This is a good topic and I'm glad you brought it up.

When you consider the light refraction coefficient of the gradients, the
first thought would be a blue grey, but clearly once you run the
numbers, you see quickly why that is wrong.

In the CRC Handbook of Chemistry & Physics, page E-224 in my 56th
Edition, I
find the following formulae:

(n-1) *10^7 = const1 + const2 / lambda^2 + const3 / lambda^4

where n is the required refractive index
lambda is the wavelength in micron (millionths of a metre -- in the actual
printed formula there are additional factors of 10^8 & 10^16 to cope with
the fact that visible spectral wavelengths are usually quoted in Aangstroem)

The values of the three constants in dry air at a pressure of 1 atm are

const1 const2 const3
30 deg C 2589.72 12.259 0.2576
15 deg C 2726.43 12.288 0.3555
0 deg C 2875.66 13.412 0.3777

A correction is given in the source to allow for air which has a water
vapour content.

To a first approximation, the variation of the refractive index with
temperature and pressure can be attributed to changes in gas density, and
the gas density can be obtained from the ideal gas equation:

(n-1) proportional to m/V = pM/RT.

But that is just a first approximation, that does not hold up for more
precise work.

Atmospheric refraction slightly increases the observed elevation angle
of a peak relative to the observer. The effect is actually quite
complicated, since it depends on the precise atmospheric conditions,
including atmospheric pressure, temperature, and water vapor content,
and thus varies with time and the altitudes of the observer and the
observed peak. Fortunately, the effect of refraction is less than ~15%
of the effect due to the curvature of the Earth, and typically only
increases the observed elevation angle by less than 0.1°.

Refraction is caused by two effects. First, light likes to travel on
the path that gets to the observer in the minimum time. (Light is,
after all, the fastest thing in the Universe, so you wouldn't expect it
would like to take a longer path than it had to, right?) The speed of
light is the speed of light in a vacuum divided by the index of
refraction. Second, the index of refraction of the atmosphere depends
on atmospheric pressure and amount of water present, which change with
height in the atmosphere. Therefore light actually travels on a curved
path in the atmosphere from one object to another. The path goes higher
than the straight-line distance in order to take advantage of the faster
speed higher in the atmosphere. Because the path is so curved, the
observer must always look a bit higher to see the light rays coming back
down from that higher elevation.

Clearly refraction must depend on some power of the distance. If you
are observing something close by, light can't get to you any quicker by
travelling very far upward. However, if you are far away from an
object, light can take advantage of the faster speed at higher elevation
and deviate more significantly from a straight line.

Astraightforward calculation gives the following formula for the angular
change with distance due to refraction between the observer at elevation
Zo (measured in km) who also is observing a peak at elevation Zo:

theta = [ 1.6 * c / { 2*(1+a) } ] * d

where d is the distance in miles, theta is in radians, and

a = 2.9e-4 * exp(-Zo/10 km) / (1 + 2.9 * To / 760)

b = 2.9 * alpha / {760 * (1 + 2.9 * To / 760) }

c = a * (b - 1/10 km)

alpha = 6.5° C. / km

The calculation assumes an atmosphere whose pressure, temperature (To in
Celsius) and water content only varies with elevation, and an atmosphere
whose temperature varies with elevation with the slope -alpha. (Alpha
is defined as the rate of temperature drop with altitude, and so is
positive in the above formulae in the normal case where the temperature
drops with altitude.) "exp" is the exponential function.

Note that this calculation assumes quite a bit. The real atmosphere can
vary markedly horizontally, can have temperature inversions, can change
its humidity, and have additional components like dust that change the
index of refraction. The observer and observed peak are not always at
the same elevation assumed in the derivation of this formula. Hence
there are no guarantees that this formula will always give accurate
results. However, on average, this formula probably gives the correct
average answer. The results of this formula at sea level are given in
surveying books as the proper term to use.

The book Elementary Surveying gives the equivalent formula in terms of
the "elevation loss" in feet of the observed object with distance:

elevation loss = 0.574 * d^2,

which is said to apply to near horizontal shots. What the book doesn't
say is that this formula is only correct near sea level.

The elevation loss formula consists of two terms:
The curvature of the Earth term, with coefficient 0.662 ( = 5280 / (2 *
R) = 5280 / (2*3986) = 0.662). The 5280 converts the final units in the
above formula to feet.
The atmospheric refraction term, which is therefore taken to have a
coefficient of 0.574 - 0.662 = -0.088.

The formula above gives a coefficient of 0.088 at an elevation of 0' and
a temperature of 65° F. However, the coefficient varies markedly with
temperature and elevation.

If you've read this far, then you also know that the density of the
water has a great deal to do with the light angle of refraction.
However, the density of the author of this question is beyond any
possible scientific measurement.


Gone Angling wrote:
I'm planning on building a cajun sneak boat. I want to use it for a fishing
situation in a small shallow bay. What is the optimal colour to paint the
submerged portion so it doesn't scare the fish ? It's a nice starter project.

Whew !!!!!
--
Jerry Barton
www.jerrys-world.com

"Fritz Nordengren" wrote in message
news:ChPqb.139951$e01.468128@attbi_s02...
This is a good topic and I'm glad you brought it up.

When you consider the light refraction coefficient of the gradients,
the
first thought would be a blue grey, but clearly once you run the
numbers, you see quickly why that is wrong.

In the CRC Handbook of Chemistry & Physics, page E-224 in my 56th
Edition, I
find the following formulae:

(n-1) *10^7 = const1 + const2 / lambda^2 + const3 / lambda^4

where n is the required refractive index
lambda is the wavelength in micron (millionths of a metre -- in the
actual
printed formula there are additional factors of 10^8 & 10^16 to cope
with
the fact that visible spectral wavelengths are usually quoted in
Aangstroem)

The values of the three constants in dry air at a pressure of 1 atm
are

const1 const2 const3
30 deg C 2589.72 12.259 0.2576
15 deg C 2726.43 12.288 0.3555
0 deg C 2875.66 13.412 0.3777

A correction is given in the source to allow for air which has a
water
vapour content.

To a first approximation, the variation of the refractive index with
temperature and pressure can be attributed to changes in gas
density, and
the gas density can be obtained from the ideal gas equation:

(n-1) proportional to m/V = pM/RT.

But that is just a first approximation, that does not hold up for
more
precise work.

Atmospheric refraction slightly increases the observed elevation
angle
of a peak relative to the observer. The effect is actually quite
complicated, since it depends on the precise atmospheric conditions,
including atmospheric pressure, temperature, and water vapor
content,
and thus varies with time and the altitudes of the observer and the
observed peak. Fortunately, the effect of refraction is less than
~15%
of the effect due to the curvature of the Earth, and typically only
increases the observed elevation angle by less than 0.1°.

Refraction is caused by two effects. First, light likes to travel
on
the path that gets to the observer in the minimum time. (Light is,
after all, the fastest thing in the Universe, so you wouldn't expect
it
would like to take a longer path than it had to, right?) The speed
of
light is the speed of light in a vacuum divided by the index of
refraction. Second, the index of refraction of the atmosphere
depends
on atmospheric pressure and amount of water present, which change
with
height in the atmosphere. Therefore light actually travels on a
curved
path in the atmosphere from one object to another. The path goes
higher
than the straight-line distance in order to take advantage of the
faster
speed higher in the atmosphere. Because the path is so curved, the
observer must always look a bit higher to see the light rays coming
back
down from that higher elevation.

Clearly refraction must depend on some power of the distance. If
you
are observing something close by, light can't get to you any quicker
by
travelling very far upward. However, if you are far away from an
object, light can take advantage of the faster speed at higher
elevation
and deviate more significantly from a straight line.

Astraightforward calculation gives the following formula for the
angular
change with distance due to refraction between the observer at
elevation
Zo (measured in km) who also is observing a peak at elevation Zo:

theta = [ 1.6 * c / { 2*(1+a) } ] * d

where d is the distance in miles, theta is in radians, and

a = 2.9e-4 * exp(-Zo/10 km) / (1 + 2.9 * To / 760)

b = 2.9 * alpha / {760 * (1 + 2.9 * To / 760) }

c = a * (b - 1/10 km)

alpha = 6.5° C. / km

The calculation assumes an atmosphere whose pressure, temperature
(To in
Celsius) and water content only varies with elevation, and an
atmosphere
whose temperature varies with elevation with the slope -alpha.
(Alpha
is defined as the rate of temperature drop with altitude, and so is
positive in the above formulae in the normal case where the
temperature
drops with altitude.) "exp" is the exponential function.

Note that this calculation assumes quite a bit. The real atmosphere
can
vary markedly horizontally, can have temperature inversions, can
change
its humidity, and have additional components like dust that change
the
index of refraction. The observer and observed peak are not always
at
the same elevation assumed in the derivation of this formula. Hence
there are no guarantees that this formula will always give accurate
results. However, on average, this formula probably gives the
correct
average answer. The results of this formula at sea level are given
in
surveying books as the proper term to use.

The book Elementary Surveying gives the equivalent formula in terms
of
the "elevation loss" in feet of the observed object with distance:

elevation loss = 0.574 * d^2,

which is said to apply to near horizontal shots. What the book
doesn't
say is that this formula is only correct near sea level.

The elevation loss formula consists of two terms:
The curvature of the Earth term, with coefficient 0.662 ( = 5280 /
(2 *
R) = 5280 / (2*3986) = 0.662). The 5280 converts the final units in
the
above formula to feet.
The atmospheric refraction term, which is therefore taken to have a
coefficient of 0.574 - 0.662 = -0.088.

The formula above gives a coefficient of 0.088 at an elevation of 0'
and
a temperature of 65° F. However, the coefficient varies markedly
with
temperature and elevation.

If you've read this far, then you also know that the density of the
water has a great deal to do with the light angle of refraction.
However, the density of the author of this question is beyond any
possible scientific measurement.


Gone Angling wrote:
I'm planning on building a cajun sneak boat. I want to use it for
a fishing
situation in a small shallow bay. What is the optimal colour to
paint the
submerged portion so it doesn't scare the fish ? It's a nice
starter project.

In the CRC Handbook of Chemistry & Physics, page E-224 in my 56th
Edition, I
find the following formulae:
(n-1) *10^7 =3D const1 + const2 / lambda^2 + const3 / lambda^4
where n is the required refractive index lambda is the wavelength in
micron ......snip.... =A0 =A0 =A0 =A0 =A0 =A0 =A0 2726.43 =A0 =A0 =A0
=A0 12.288 =A0 =A0 =A0 =A0 0.3555
0 deg C =A0 =A0 =A0 =A0 2875.66 =A0 =A0 =A0 =A0 13.412 =A0 =A0 =A0 =A0
0.3777

Huh? Lmao

LMAO
"J Buck" wrote in message
...
In the CRC Handbook of Chemistry & Physics, page E-224 in my 56th
Edition, I
find the following formulae:
(n-1) *10^7 = const1 + const2 / lambda^2 + const3 / lambda^4
where n is the required refractive index lambda is the wavelength in
micron ......snip.... 2726.43
12.288 0.3555
0 deg C 2875.66 13.412
0.3777

Huh? Lmao

Fritz Nordengren wrote:

This is a good topic and I'm glad you brought it up.

When you consider the light refraction coefficient of the gradients, the
first thought would be a blue grey, but clearly once you run the
numbers, you see quickly why that is wrong. analysis snipped

So it would be sky blue then ? ;-)

--
Ken Fortenberry

Aint sure what I just read but it sounds like the motto: If ya can't dazzle
em with brilliance, baffle em with bull****. I might as well been trying
to read Aramic.

"Fritz Nordengren" wrote in message
news:ChPqb.139951$e01.468128@attbi_s02...
This is a good topic and I'm glad you brought it up.

When you consider the light refraction coefficient of the gradients, the
first thought would be a blue grey, but clearly once you run the
numbers, you see quickly why that is wrong.

In the CRC Handbook of Chemistry & Physics, page E-224 in my 56th
Edition, I
find the following formulae:

(n-1) *10^7 = const1 + const2 / lambda^2 + const3 / lambda^4

where n is the required refractive index
lambda is the wavelength in micron (millionths of a metre -- in the actual
printed formula there are additional factors of 10^8 & 10^16 to cope with
the fact that visible spectral wavelengths are usually quoted in
Aangstroem)

The values of the three constants in dry air at a pressure of 1 atm are

const1 const2 const3
30 deg C 2589.72 12.259 0.2576
15 deg C 2726.43 12.288 0.3555
0 deg C 2875.66 13.412 0.3777

A correction is given in the source to allow for air which has a water
vapour content.

To a first approximation, the variation of the refractive index with
temperature and pressure can be attributed to changes in gas density, and
the gas density can be obtained from the ideal gas equation:

(n-1) proportional to m/V = pM/RT.

But that is just a first approximation, that does not hold up for more
precise work.

Atmospheric refraction slightly increases the observed elevation angle
of a peak relative to the observer. The effect is actually quite
complicated, since it depends on the precise atmospheric conditions,
including atmospheric pressure, temperature, and water vapor content,
and thus varies with time and the altitudes of the observer and the
observed peak. Fortunately, the effect of refraction is less than ~15%
of the effect due to the curvature of the Earth, and typically only
increases the observed elevation angle by less than 0.1°.

Refraction is caused by two effects. First, light likes to travel on
the path that gets to the observer in the minimum time. (Light is,
after all, the fastest thing in the Universe, so you wouldn't expect it
would like to take a longer path than it had to, right?) The speed of
light is the speed of light in a vacuum divided by the index of
refraction. Second, the index of refraction of the atmosphere depends
on atmospheric pressure and amount of water present, which change with
height in the atmosphere. Therefore light actually travels on a curved
path in the atmosphere from one object to another. The path goes higher
than the straight-line distance in order to take advantage of the faster
speed higher in the atmosphere. Because the path is so curved, the
observer must always look a bit higher to see the light rays coming back
down from that higher elevation.

Clearly refraction must depend on some power of the distance. If you
are observing something close by, light can't get to you any quicker by
travelling very far upward. However, if you are far away from an
object, light can take advantage of the faster speed at higher elevation
and deviate more significantly from a straight line.

Astraightforward calculation gives the following formula for the angular
change with distance due to refraction between the observer at elevation
Zo (measured in km) who also is observing a peak at elevation Zo:

theta = [ 1.6 * c / { 2*(1+a) } ] * d

where d is the distance in miles, theta is in radians, and

a = 2.9e-4 * exp(-Zo/10 km) / (1 + 2.9 * To / 760)

b = 2.9 * alpha / {760 * (1 + 2.9 * To / 760) }

c = a * (b - 1/10 km)

alpha = 6.5° C. / km

The calculation assumes an atmosphere whose pressure, temperature (To in
Celsius) and water content only varies with elevation, and an atmosphere
whose temperature varies with elevation with the slope -alpha. (Alpha
is defined as the rate of temperature drop with altitude, and so is
positive in the above formulae in the normal case where the temperature
drops with altitude.) "exp" is the exponential function.

Note that this calculation assumes quite a bit. The real atmosphere can
vary markedly horizontally, can have temperature inversions, can change
its humidity, and have additional components like dust that change the
index of refraction. The observer and observed peak are not always at
the same elevation assumed in the derivation of this formula. Hence
there are no guarantees that this formula will always give accurate
results. However, on average, this formula probably gives the correct
average answer. The results of this formula at sea level are given in
surveying books as the proper term to use.

The book Elementary Surveying gives the equivalent formula in terms of
the "elevation loss" in feet of the observed object with distance:

elevation loss = 0.574 * d^2,

which is said to apply to near horizontal shots. What the book doesn't
say is that this formula is only correct near sea level.

The elevation loss formula consists of two terms:
The curvature of the Earth term, with coefficient 0.662 ( = 5280 / (2 *
R) = 5280 / (2*3986) = 0.662). The 5280 converts the final units in the
above formula to feet.
The atmospheric refraction term, which is therefore taken to have a
coefficient of 0.574 - 0.662 = -0.088.

The formula above gives a coefficient of 0.088 at an elevation of 0' and
a temperature of 65° F. However, the coefficient varies markedly with
temperature and elevation.

If you've read this far, then you also know that the density of the
water has a great deal to do with the light angle of refraction.
However, the density of the author of this question is beyond any
possible scientific measurement.


Gone Angling wrote:
I'm planning on building a cajun sneak boat. I want to use it for a
fishing
situation in a small shallow bay. What is the optimal colour to paint
the
submerged portion so it doesn't scare the fish ? It's a nice starter
project.

Melt fluorocarbon line in a glue gun and smear it all over da boat. Fish
can't see ya commin dat way. Optimum water refectory color dats what
fluorocarbon can do for ya.


"Gone Angling" wrote in message
...
I'm planning on building a cajun sneak boat. I want to use it for a
fishing
situation in a small shallow bay. What is the optimal colour to paint the
submerged portion so it doesn't scare the fish ? It's a nice starter
project.