Hi.
I have a home brew laser. 1500x1000mm bed. It’s an 80W CO2.
Something I am noticing is that my cutting and engraving power lessens the further away from the tube I am.
i.e. When working closer to the tube output I can engrave maybe 0.25mm in to the wood, however at a distance I am just discoloring the wood,
Is this normal and do people factor this in when working at the further edges of their work.
I am very sure I have the laser mirrors aligned correctly and they are cleaned regularly.
With CO2 laser, it is “normal” to have some scatter before the beam hits the lens. However, the probability that your laser beam is not parallel and 100% centered is high and has a greater impact. Up to 2% energy loss per mirror and optics in a laser system is further calculated. With theoretical 2600mm you also have a long way to go before the laser hits the lens at the rear end. How big is your optics in diameter?
Hi.
Can you be specific re “How Big are your Optics”… Mirror, Lens, etc.
But - If I was to calculate the full length of the bean then I would be getting close to 2500mm to the furthest corner
Yes, I mean the diameter of the mirrors and lens. At that total length, I have taken the last cm, all the way up to the exit from the tube in the accounts. +/- a few cm
There is some beam divergence, but it is a much longer distance than normal light. The inverse square law Does Apply, but with a longer distance involved. If there were no beam divergence, then this law wouldn’t apply.
Charlie
According to one of the radiology sites explaining how the inverse square law works.
They also point out…
A situation where the inverse square law does not apply is in the case of a laser beam. Because the light is essentially parallel, the intensity is constant and does not depend on distance from the source.
We all know that energy does ‘spread out’ over distance. I also know when I transmit a signal to my antenna the power is ‘constant’ yet the inverse law still holds.
They have been shooting lasers at the moon since the 70’s. The following is from NASA… it’s about laser beams reflected from the moon. They use infra red because it passes through the atmosphere with less loss and dispersion.
light beam that’s about 10 feet, or a few meters, wide on the ground can spread out to more than 1 mile, or 2 kilometers, by the time it reaches the Moon’s surface, and much wider when it bounces back
I’ve seen photographs of these firing, obviously not IR, but they don’t look 3 meters in diameter…
I have a 5030 machine, with home and the beam entry at the rear/left, The beam is smaller at the tube than at the front right side, the maximum distance from the tube.
Keep in mind that a 20mm diameter mirror is on a 45 deg angle, so that axis isn’t 20mm target it’s a 14 mm target. A 8mm beam uses 14mm - 8mm or 6mm of the mirror, giving you only 3mm on each side for error. An 18mm has only 12mm - 8 or 4mm or only 2mm each side for ‘wiggle’ room.
This site is for lasers, but I really couldn’t follow the math this morning, but it doesn’t look like an inverse square law either.
The simple fact of what makes a laser useful is the lens. If the whole beam is properly aligned, there is very little energy loss across the whole beam in that short of a distance, although the beam is larger. It’s the lens. If it’s isn’t making it though the optical path cleanly, that’s a problem.
Russ Sadler has also just made a new video on the subject, it’s interesting as always.
I can’t add any facts to the title of this thread but I can point out that a machine that large would be difficult to align and keep in alignment both during the cut and over time. Because of vibration and any warpage in the rail plane.
Is it possible that what looks good during your static alignment process changes at cutting speeds?
I read this, also and it does still apply, the beam of a laser, by the time it travels to the moon, is a Very Large spot lol. This is why the reflectors can work, being small. A red dot laser does spread, too. If you aim it at a distant target, 10 miles away, the beam looks from your standpoint as a very small dot, but at the target, it is 6 feet high.
Charlie