Aluminum foil

Ceebus

Active Member
Is aluminum foil any good help? Does it really do the trick to reflect the lights?

What about mirrors? Anybody tried that?
A couple people have said that aluminum foil doesnt reflect much and what it does reflect can possibly burn holes in your leaves.
 

pennywise619

Well-Known Member
The cheapest shit that actually works in an emergency blanket from walmart almost identical to mylar, but with creases from the packaging. They sell them for $1.98. Also, if your growing with cfls aluminum foil will not give you hot spots, but is not recommended for HID lamps.
 

hootie233

Well-Known Member
If i were you, i would go to your auto zone or checker auto parts and you can get a mylar winsheild thingy for 8 bucks tops and its more than big enough and its ridged enough to stand up on its own.
 

pennywise619

Well-Known Member
Isnt mylar the same film used on printed balloons? The ones that normally look like oversized M&Ms

Yes, are you planning on using a bunch of ballon, LOL.... Dude just buy an emergegnecy blaket from walmart $1.98 in the camping section. This will be a well spent $1.98....
 

Brick Top

New Member
Is aluminum foil any good help? Does it really do the trick to reflect the lights?

What about mirrors? Anybody tried that?

Aluminum foil will at best reflect 50% to 55% of the light rays and that will decrease with any crinkling and or damage to the aluminum foil. Any unevenness/crinkling of the aluminum foil can and will create hot spots. You may get lucky and some or most will be off to the sides of your plants and not damage them but any that hits your plants will be a problem.

Flat white paint will reflect 80% to 85% of light rays. Mylar is best and will reflect around 90% to 92% of light rays but it is not inexpensive and if you try to save money and buy the thinner Mylar it is fragile and easily torn and like with aluminum foil any damage will decrease its reflectivity.

The easiest least expensive and yet highly efficient thing to use for light reflection is flat white paint.

You asked about using mirrors, well that is a matter of geometry. A perfect mirror reflects all of the light that strikes it, but reflects it as regular, or specular, reflection. That means all the light is reflected at an angle equal to the angle of incidence and opposite the normal to the surface. A perfect white also reflects all the light incident on it, but reflects that light diffusely. In other words, the white scatters the incident light in all directions. That's why a white object looks white from all angles, but you can only see yourself in a mirror when you look at it directly (in line with the normal to it's surface).
 

SlickRickMcgee

Well-Known Member
Aluminum foil will at best reflect 50% to 55% of the light rays and that will decrease with any crinkling and or damage to the aluminum foil. Any unevenness/crinkling of the aluminum foil can and will create hot spots. You may get lucky and some or most will be off to the sides of your plants and not damage them but any that hits your plants will be a problem.

Flat white paint will reflect 80% to 85% of light rays. Mylar is best and will reflect around 90% to 92% of light rays but it is not inexpensive and if you try to save money and buy the thinner Mylar it is fragile and easily torn and like with aluminum foil any damage will decrease its reflectivity.

The easiest least expensive and yet highly efficient thing to use for light reflection is flat white paint.

You asked about using mirrors, well that is a matter of geometry. A perfect mirror reflects all of the light that strikes it, but reflects it as regular, or specular, reflection. That means all the light is reflected at an angle equal to the angle of incidence and opposite the normal to the surface. A perfect white also reflects all the light incident on it, but reflects that light diffusely. In other words, the white scatters the incident light in all directions. That's why a white object looks white from all angles, but you can only see yourself in a mirror when you look at it directly (in line with the normal to it's surface).
that's great info. thanks!
 
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