Going from HPS to LED. Worth it?

Dankistino

Well-Known Member
Then educate me as to why the laws of physics don't apply to cobs?
"Cobs lack initial value" what? Fucking incoherent jackass.

#1 "penetration" is nothing more than par
#2 inverse sq law does not account for reflectivity
#3 hps emit light at 360°, this structural shortcoming places bulbs mounted above canopy at a huge disadvantage.

You are far too stupid and confused to be arguing. You need to stfu and get a grip. You are boldly shouting out of your asshole.
 

BecauseIgotHigh

Well-Known Member
lol I was gonna tell you to scrog :) I scrog my leds , found out they yield a lot more like this about 2 years ago, they lack the penetration but if you can fill out a screen you can get the weight.
hahaha yodaweed, i had to scrog all 3 of my previous grow because the COBs penetration was not cutting it for me. That's why many cobs user do scrog, because COBs and all cob users know they use it because we can spread out the lighting.

You are crazy if you think COBs can't penetrate, and you are basing that idea on literally nothing.
 

Olive Drab Green

Well-Known Member
hahaha yodaweed, i had to scrog all 3 of my previous grow because the COBs penetration was not cutting it for me. That's why many cobs user do scrog, because COBs and all cob users know they use it because we can spread out the lighting.
If you don't have penetration, I wonder how low your amperage is?
 

BecauseIgotHigh

Well-Known Member
When was the last time you fools used LEDs? When 3 watt diodes were the new thing? If Yoda's saying 4 years, I have a feeling his experience has been with panels or old COBs. I could be wrong, though.
lol didn't get sucked in that hype. i guess im not a fool, since i've been using cxb for about a year now. i still got the cxa3590 5000k at the back somewhere. Now that's a great efficient light for vegging there lads.

If you don't have penetration, I wonder how low your amperage is?
72v 700ma
 

Dankistino

Well-Known Member
lol didn't get sucked in that hype. i guess im not a fool, since i've been using cxb for about a year now. i still got the cxa3590 5000k at the back somewhere. Now that's a great efficient light for vegging there lads.



72v 700ma
Bro plenty of people have 20-24 inch bud filled stems using tbe exact same mA. Your lack of skill isnt an explanation of failed penetration.
 

Olive Drab Green

Well-Known Member
lol didn't get sucked in that hype. i guess im not a fool, since i've been using cxb for about a year now. i still got the cxa3590 5000k at the back somewhere. Now that's a great efficient light for vegging there lads.



72v 700ma
That explains it. Your intensity was shit. I run 2 36v 3500k CXB3590 top bins on their own 2150mA drivers. I should have a Vero29C 3500k running 1670? mA to add by tomorrow.

You were running low amperage, and using an older chip in a less-than-ideal color spectrum. And running a single obsolete light for what kind of results, I wonder? A single medium plant?
 
Last edited:

Yodaweed

Well-Known Member
"Cobs lack initial value" what? Fucking incoherent jackass.

#1 "penetration" is nothing more than par
#2 inverse sq law does not account for reflectivity
#3 hps emit light at 360°, this structural shortcoming places bulbs mounted above canopy at a huge disadvantage.

You are far too stupid and confused to be arguing. You need to stfu and get a grip. You are boldly shouting out of your asshole.
The intensity (or illuminance or irradiance) of light or other linear waves radiating from a point source (energy per unit of area perpendicular to the source) is inversely proportional to the square of the distance from the source; so an object (of the same size) twice as far away, receives only one-quarter the energy (in the same time period).

More generally, the irradiance, i.e., the intensity (or power per unit area in the direction of propagation), of a spherical wavefront varies inversely with the square of the distance from the source (assuming there are no losses caused by absorption or scattering).

For example, the intensity of radiation from the Sun is 9126 watts per square meter at the distance of Mercury (0.387 AU); but only 1367 watts per square meter at the distance of Earth (1 AU)—an approximate threefold increase in distance results in an approximate ninefold decrease in intensity of radiation.

For non-isotropic radiators such as parabolic antennas, headlights, and lasers, the effective origin is located far behind the beam aperture. If you are close to the origin, you don't have to go far to double the radius, so the signal drops quickly. When you are far from the origin and still have a strong signal, like with a laser, you have to travel very far to double the radius and reduce the signal. This means you have a stronger signal or have antenna gain in the direction of the narrow beam relative to a wide beam in all directions of an isotropic antenna.

In photography and stage lighting, the inverse-square law is used to determine the "fall off" or the difference in illumination on a subject as it moves closer to or further from the light source. For quick approximations, it is enough to remember that doubling the distance reduces illumination to one quarter;[8] or similarly, to halve the illumination increase the distance by a factor of 1.4 (the square root of 2), and to double illumination, reduce the distance to 0.7 (square root of 1/2). When the illuminant is not a point source, the inverse square rule is often still a useful approximation; when the size of the light source is less than one-fifth of the distance to the subject, the calculation error is less than 1%.[9]

The fractional reduction in electromagnetic fluence (Φ) for indirectly ionizing radiation with increasing distance from a point source can be calculated using the inverse-square law. Since emissions from a point source have radial directions, they intercept at a perpendicular incidence. The area of such a shell is 4πr 2 where r is the radial distance from the center. The law is particularly important in diagnostic radiography and radiotherapy treatment planning, though this proportionality does not hold in practical situations unless source dimensions are much smaller than the distance. As stated in fourier theory of heat "as the point source is magnification by distances , its radiation is dilute proportional to the sin of the angle, of the increasing circumference arc from the point of origin"

Example[edit]
Let the total power radiated from a point source, for example, an omnidirectional isotropic radiator, be P. At large distances from the source (compared to the size of the source), this power is distributed over larger and larger spherical surfaces as the distance from the source increases. Since the surface area of a sphere of radius r is A = 4πr 2, then intensity I (power per unit area) of radiation at distance r is

I = P A = P 4 π r 2 . {\displaystyle I={\frac {P}{A}}={\frac {P}{4\pi r^{2}}}.\,}

The energy or intensity decreases (divided by 4) as the distance r is doubled; measured in dB it would decrease by 6.02 dB per doubling of distance.
 

Olive Drab Green

Well-Known Member
The intensity (or illuminance or irradiance) of light or other linear waves radiating from a point source (energy per unit of area perpendicular to the source) is inversely proportional to the square of the distance from the source; so an object (of the same size) twice as far away, receives only one-quarter the energy (in the same time period).

More generally, the irradiance, i.e., the intensity (or power per unit area in the direction of propagation), of a spherical wavefront varies inversely with the square of the distance from the source (assuming there are no losses caused by absorption or scattering).

For example, the intensity of radiation from the Sun is 9126 watts per square meter at the distance of Mercury (0.387 AU); but only 1367 watts per square meter at the distance of Earth (1 AU)—an approximate threefold increase in distance results in an approximate ninefold decrease in intensity of radiation.

For non-isotropic radiators such as parabolic antennas, headlights, and lasers, the effective origin is located far behind the beam aperture. If you are close to the origin, you don't have to go far to double the radius, so the signal drops quickly. When you are far from the origin and still have a strong signal, like with a laser, you have to travel very far to double the radius and reduce the signal. This means you have a stronger signal or have antenna gain in the direction of the narrow beam relative to a wide beam in all directions of an isotropic antenna.

In photography and stage lighting, the inverse-square law is used to determine the "fall off" or the difference in illumination on a subject as it moves closer to or further from the light source. For quick approximations, it is enough to remember that doubling the distance reduces illumination to one quarter;[8] or similarly, to halve the illumination increase the distance by a factor of 1.4 (the square root of 2), and to double illumination, reduce the distance to 0.7 (square root of 1/2). When the illuminant is not a point source, the inverse square rule is often still a useful approximation; when the size of the light source is less than one-fifth of the distance to the subject, the calculation error is less than 1%.[9]

The fractional reduction in electromagnetic fluence (Φ) for indirectly ionizing radiation with increasing distance from a point source can be calculated using the inverse-square law. Since emissions from a point source have radial directions, they intercept at a perpendicular incidence. The area of such a shell is 4πr 2 where r is the radial distance from the center. The law is particularly important in diagnostic radiography and radiotherapy treatment planning, though this proportionality does not hold in practical situations unless source dimensions are much smaller than the distance. As stated in fourier theory of heat "as the point source is magnification by distances , its radiation is dilute proportional to the sin of the angle, of the increasing circumference arc from the point of origin"

Example[edit]
Let the total power radiated from a point source, for example, an omnidirectional isotropic radiator, be P. At large distances from the source (compared to the size of the source), this power is distributed over larger and larger spherical surfaces as the distance from the source increases. Since the surface area of a sphere of radius r is A = 4πr 2, then intensity I (power per unit area) of radiation at distance r is

I = P A = P 4 π r 2 . {\displaystyle I={\frac {P}{A}}={\frac {P}{4\pi r^{2}}}.\,}

The energy or intensity decreases (divided by 4) as the distance r is doubled; measured in dB it would decrease by 6.02 dB per doubling of distance.
So, Yoda, you said 4 years of LED experience. What light, what specs.
 

Dankistino

Well-Known Member
The intensity (or illuminance or irradiance) of light or other linear waves radiating from a point source (energy per unit of area perpendicular to the source) is inversely proportional to the square of the distance from the source; so an object (of the same size) twice as far away, receives only one-quarter the energy (in the same time period).

More generally, the irradiance, i.e., the intensity (or power per unit area in the direction of propagation), of a spherical wavefront varies inversely with the square of the distance from the source (assuming there are no losses caused by absorption or scattering).

For example, the intensity of radiation from the Sun is 9126 watts per square meter at the distance of Mercury (0.387 AU); but only 1367 watts per square meter at the distance of Earth (1 AU)—an approximate threefold increase in distance results in an approximate ninefold decrease in intensity of radiation.

For non-isotropic radiators such as parabolic antennas, headlights, and lasers, the effective origin is located far behind the beam aperture. If you are close to the origin, you don't have to go far to double the radius, so the signal drops quickly. When you are far from the origin and still have a strong signal, like with a laser, you have to travel very far to double the radius and reduce the signal. This means you have a stronger signal or have antenna gain in the direction of the narrow beam relative to a wide beam in all directions of an isotropic antenna.

In photography and stage lighting, the inverse-square law is used to determine the "fall off" or the difference in illumination on a subject as it moves closer to or further from the light source. For quick approximations, it is enough to remember that doubling the distance reduces illumination to one quarter;[8] or similarly, to halve the illumination increase the distance by a factor of 1.4 (the square root of 2), and to double illumination, reduce the distance to 0.7 (square root of 1/2). When the illuminant is not a point source, the inverse square rule is often still a useful approximation; when the size of the light source is less than one-fifth of the distance to the subject, the calculation error is less than 1%.[9]

The fractional reduction in electromagnetic fluence (Φ) for indirectly ionizing radiation with increasing distance from a point source can be calculated using the inverse-square law. Since emissions from a point source have radial directions, they intercept at a perpendicular incidence. The area of such a shell is 4πr 2 where r is the radial distance from the center. The law is particularly important in diagnostic radiography and radiotherapy treatment planning, though this proportionality does not hold in practical situations unless source dimensions are much smaller than the distance. As stated in fourier theory of heat "as the point source is magnification by distances , its radiation is dilute proportional to the sin of the angle, of the increasing circumference arc from the point of origin"

Example[edit]
Let the total power radiated from a point source, for example, an omnidirectional isotropic radiator, be P. At large distances from the source (compared to the size of the source), this power is distributed over larger and larger spherical surfaces as the distance from the source increases. Since the surface area of a sphere of radius r is A = 4πr 2, then intensity I (power per unit area) of radiation at distance r is

I = P A = P 4 π r 2 . {\displaystyle I={\frac {P}{A}}={\frac {P}{4\pi r^{2}}}.\,}

The energy or intensity decreases (divided by 4) as the distance r is doubled; measured in dB it would decrease by 6.02 dB per doubling of distance.
Youre a jack ass.
 

Olive Drab Green

Well-Known Member
@Yodaweed I just looked at your blog. You use fucking blurple lights. You don't even use COBs. And if you do, what specs? Because if you have never used COBs, you really do need to shut the fuck up, because COBs are WAY different than panel LEDs. And your stubbornness to understand is solely a result of bias and wanting to bend science and reality to fit your understanding of it.
 

BecauseIgotHigh

Well-Known Member
That explains it. Your intensity was shit. I run 2 36v 3500k CXB3590 top bins on their own 2150mA drivers. I should have a Vero29C 3500k running 1670? mA to add by tomorrow.

You were running low amperage, and using an older chip in a less-than-ideal color spectrum. And running a single obsolete light for what kind of results, I wonder? A single medium plant?
3000k cxb3590 72v is still a top bin to me. 8 chips too? how many chips are you running in a small area. obsolete???? you know whats obsolete??? Motorola is obsolete. DDR2 is becoming obsolete. Now the argument is my intensity isn't bright enough, pump up the amps... fuck that, wasn't the point of going 700ma was to get more efficiency? Now, you're saying run it high, maybe towards 1000ma. now efficiency doesn't mean squat.

A single medium plant? lol you sir have no clue
Fuck this thread, to you sir stop jacking people thread and throwing them info, without letting them decide what's right for them.
You have no clue if they're doing a big grow op like factory size. then ill be looking towards getting hps to do the job.
where the fuck is the OP anyways?
 

Olive Drab Green

Well-Known Member
3000k cxb 72v is still a top bin to me. 8 chips too? how many chips are you running in a small area. obsolete???? you know whats obsolete??? Motorola is obsolete. DDR2 is becoming obsolete. Now the argument is my intensity isn't bright enough, pump up the amps... fuck that, wasn't the point of going 700ma was to get more efficiency? Now, you're saying run it high, maybe towards 1000ma. now efficiency doesn't mean squat.

A single medium plant? lol you sir have no clue
Fuck this thread, to you sir stop jacking people thread and throwing them info, without letting them decide what's right for them.
You have no clue if they're doing a big grow op like factory size. then ill be looking towards getting hps to do the job.
where the fuck is the OP anyways?
You used a 5000k CXA3590 and you ran it about 50w. I run mine about 78w, and the last one about 85w. Your kelvin and amperage were the biggest issue. Probably should have run it at 1-1.1 amps, at least.

Efficiency still means plenty. With 274w, I'm getting the PPFD of a 600w HPS.
 
Top