First reflection equation?

[SouthSIDE Glen]

First off, my apologies if this question has been answered a million times, but I don't frequent this forum, and several tries at an "advanced search" in the lousy search engine that vBulletin has has not yielded a good result. The question itself is simple...

Assuming one does not have the ability to use a mirror to find the first reflection points on the walls and ceiling, what would be the mathematical equation for figuring it out.

I know how to do it for two points that are the same distance from the wall, but not for two points that are of differeng distances (such as a nearfield monitor and a human head ).

TIA as always,

G.

 

[suprstar]

The angles between monitor M and reflection point R, and reflection point and head H (MRX and HRX) must be equal, thats what defines a reflection point. Those lines can be any arbitrary length, thats why the equation is simple if lines RM and RH are equal length, you know one leg's length based on the length of the other leg.

MW - distance from M to wall
MX - distance from M to line RX
HW - distance from H to wall
HX - distance from H to line RX

as long as HX/HW = MX/MW, it is a reflection point. Since there are infinite point along either line, you need more info to determine where exactly on those 2 lines the points H and M are. You need one's absolute position, and one leg of the triangle on the other side. Positions must be relative to wall and RX. There are many equations depending on what info you have, certain info is required..

 

[SouthSIDE Glen]

You need one's absolute position, and one leg of the triangle on the other side. Positions must be relative to wall and RX. There are many equations depending on what info you have, certain info is required..

I would think/hope that it could be calculated knowing just the x/y coordinates of H and M within the room.

It's incredibly easy if the distances from the wall are the same, because all one needs to know is the x coordinates of H and M. Since the distance from the wall will be the same, MX and HX will be the same length. This means that the reflection point will be exactly midway between the head and the monitor. So all one needs to do is add the x coordinate of M to the x coordinate of H and then divide that sum by 2, and we have the x coordinate for R.

When the distance to W varies, however, MX and HX will vary in length as well, so the simple average will no longer work. Its a change in the y coordinates of M or H that is causing a shift in the x position of R. This relationship is direct and linear. I just can't figure it out (or find it on the 'net .) I'dve sworn that a couple of weeks ago I was here and I tripped across an old thread where either Mr. Gervais or someone else here gave a simple formula for finding the reflection point, but I'll be damned if I can find that thread again.

And of course just when I need them, both of my local Borders happen to be out of stock on the good studio building books; there must be a run on people building home studios around here. I'll have to try some "look inside" searches at Amazon...

G.

 

[SouthSIDE Glen]

Update

Found it. Wound up dropping the acoustics angle (pun intended) and searching on calculating billiards angles . Got this right away:

http://www.jimloy.com/billiard/kick.htm

G.

 

[Ethan Winer]

Assuming one does not have the ability to use a mirror to find the first reflection points on the walls and ceiling, what would be the mathematical equation for figuring it out.

Here's mine:

How to set up a room

See Figure 4.

--Ethan

 

[SouthSIDE Glen]

Here's mine:

How to set up a room

See Figure 4.

--Ethan

Thanks, Ethan, I was hoping you might find this thread . I did check out your site, but I somehow overlooked that article.

Yeah, your formula looks basically the same as the one on the bankshot page; the only difference - which really is not a difference - is they are calculating from the cue ball (tweeter) instead from from the 3 ball (listener). But it remains basically the same equation, just plugging in a different side of the values. Your explanation and graphic of the equation are IMHO much easier to readily understand, however.

On that point, one teeny tiny suggestion? I had to refer to the graphic to see that you were in fact calculating the distance from the listener to the reflection point, and not from the tweeter to the reflection point. In the text column where you describe the equation, you say:

Let's say the speaker is 2 feet from the side wall (x1), and you're 6 feet from that same wall (x2), and the loudspeaker's tweeter is 5 feet in front of you . The distance to the center of the panel is solved as follows:​

This does not say from which side the distance is calculated, the speaker or the listener. It's not unless or until one looks at Fig. 4 that it is graphically displayed that you mean from the listener. Have you considered perhaps just adding "from the listener" to that last sentence in the text? No biggie; if I could figure it out, probably most others could as well. But if someone were to just take your text description w/o popping up the graphic, they could make a mistake.

Just a small, picky thing. Thanks for the info - here and on your high quality website in general!

G.

 

[suprstar]

They're similar triangles, so if you know the distances of lines MW and HW, they are proportional to the distances of lines MX and HX. That's all you need to know!

 

[SouthSIDE Glen]

They're similar triangles, so if you know the distances of lines MW and HW, they are proportional to the distances of lines MX and HX. That's all you need to know!

Maybe I'm misunderstanding what you're saying, but you need to know more than just MW and HW, which are essentially just the y coordinates of the two locations. The x coordinates are required as well.

But not only is that not an issue, that is exactly the problem I described; knowing only the x/y of the monitor and the head (and defining the wall as y=0), the position can be determined using the formula. Ethan takes it y difference from the listener, that billiards page as y distance from the monitor, but it's really the same thing, just deciding which triangle you want to measure.

G.

 

[Ethan Winer]

Your explanation and graphic of the equation are IMHO much easier to readily understand, however.

Yep, that's why I posted the link. Also, your clarification idea is excellent, so I just changed that last sentence to read:

"The distance between the wall opposite your head and the center of the panel is solved as follows:"​

I started to write "distance from your head" but that would imply a straight line from your head to the panel's center which is wrong. Hopefully this is clear enough?

--Ethan

 

[SouthSIDE Glen]

I started to write "distance from your head" but that would imply a straight line from your head to the panel's center which is wrong. Hopefully this is clear enough?

Good catch. Yeah, that should be clear enough, I should think.

G.