Jim Hans Posted December 19, 2013 Share Posted December 19, 2013 To all my smart decorator friends or those with smart kids.... 25 years ago I could have figured this out myself but now, well, it's easier to ask. I have 3 columns in front of my house that I want to put Christmas lights on. They are 18 feet tall and 9" in diameter. I want to spiral lights around them using 2 inch spacing. How many feet of lights will I need per column? And if you're really smart... how many 100, 70 and/or 50 bulb strings will I need for each column??? I'm sure there is a simple formula somewhere but...... Thanks in advance. Link to comment Share on other sites More sharing options...
Brian Mitchell Posted December 19, 2013 Share Posted December 19, 2013 pie r squared = about 64 inch circumference 6 wraps per foot X 18 = 108 X 64 divided by 12 inches per foot = 576 ft per column. About 24 ft per string = 24 strings per column Link to comment Share on other sites More sharing options...
wbaker4 Posted December 19, 2013 Share Posted December 19, 2013 I am not a math wiz... but a rough calculation.... Circumference is 3.1415 * diameter Circumference = 28.27 inches. 18 ft = 216 inches Length for 1 column = (216/2) * 28.27 = 3053 inches or 254 ft. per column Since you will be spiraling up you will need a little bit more to compensate for the angle. Link to comment Share on other sites More sharing options...
Brian Mitchell Posted December 19, 2013 Share Posted December 19, 2013 One of us is wrong. It's probably me. Link to comment Share on other sites More sharing options...
wbaker4 Posted December 19, 2013 Share Posted December 19, 2013 I am sure others will answer it as well. We will see! Link to comment Share on other sites More sharing options...
Brian Mitchell Posted December 19, 2013 Share Posted December 19, 2013 ya my formula was for area not circumference. Link to comment Share on other sites More sharing options...
Brian Mitchell Posted December 19, 2013 Share Posted December 19, 2013 So refiguring, it's about 11 strings per column. Link to comment Share on other sites More sharing options...
Jim Hans Posted December 19, 2013 Author Share Posted December 19, 2013 One of us is wrong. It's probably me. Pi*r squared is for the area of a circle. I am not a math wiz... but a rough calculation.... Circumference is 3.1415 * diameter Circumference = 28.27 inches. 18 ft = 216 inches Length for 1 column = (216/2) * 28.27 = 3053 inches or 254 ft. per column Since you will be spiraling up you will need a little bit more to compensate for the angle. And that is the heart of the question... how much more because of the spiraling? I figured the circumference easy enough but it isthe spiraling up calculation that I am looking for. Link to comment Share on other sites More sharing options...
DSE Posted December 19, 2013 Share Posted December 19, 2013 Spiraling or not you will be covering the same square inches Link to comment Share on other sites More sharing options...
-klb- Posted December 19, 2013 Share Posted December 19, 2013 Sqrt(28.27^2+2^2)=28.34 per turn. Link to comment Share on other sites More sharing options...
-klb- Posted December 19, 2013 Share Posted December 19, 2013 You probably also need to account for the light string not bending at the surface of the column, but somewhere within the thickness of the strand of lights. I would guess that half the average diameter of the strand would do it. Maybe repeat the calculations as if the column was 1/8 inch larger? Link to comment Share on other sites More sharing options...
-klb- Posted December 19, 2013 Share Posted December 19, 2013 You probably also need to account for the light string not bending at the surface of the column, but somewhere within the thickness of the strand of lights. I would guess that half the average diameter of the strand would do it. Maybe repeat the calculations as if the column was 1/8 inch larger? Link to comment Share on other sites More sharing options...
Jim Hans Posted December 19, 2013 Author Share Posted December 19, 2013 Sqrt(28.27^2+2^2)=28.34 per turn. Thanks! To make sure I understand the formula... if I want to do it with 3" spacing the formula would be Sqrt(circumference^2+3^2)= Yes??? Link to comment Share on other sites More sharing options...
wbaker4 Posted December 19, 2013 Share Posted December 19, 2013 Sqrt(circumference^2 + SPACING^2) * (216 / SPACING) SPACING could be 2" or 3", etc. Link to comment Share on other sites More sharing options...
Brian Mitchell Posted December 19, 2013 Share Posted December 19, 2013 2" spacing would take 108 wraps, 3" spacing will take 72 wraps or about 168 ft.String lengths vary greatly from vendor to vendor so you would just have to pick one and do the math. Going to 4" spacing brings you down to 54 wraps or about 126 ft. Link to comment Share on other sites More sharing options...
MikeH Posted December 20, 2013 Share Posted December 20, 2013 Seems to me the math will take longer than to just start wrapping lights. When you get to the last one, count the number of bulbs used and get the appropriate length you need. MikeH Link to comment Share on other sites More sharing options...
Jim Hans Posted December 20, 2013 Author Share Posted December 20, 2013 Thanks... I have the formula now and it is pretty easy to figure it out. The nice thing about the formula is I can solve for spacing based on the total length of lights I will be using. Now that I know how many more lights I will need to do this I am patiently waiting for the sales!! Link to comment Share on other sites More sharing options...
Mwhite7097 Posted December 20, 2013 Share Posted December 20, 2013 I have columns as well and found it easier to hang the light strings vertically in front of the columns. This is the part people see anyway. You will use less strings and have more light towards the front, easy to figure how many strings, and easier setup than spiraling around. Looks the same from the road. Link to comment Share on other sites More sharing options...
Jim Hans Posted December 20, 2013 Author Share Posted December 20, 2013 I have columns as well and found it easier to hang the light strings vertically in front of the columns. This is the part people see anyway. You will use less strings and have more light towards the front, easy to figure how many strings, and easier setup than spiraling around. Looks the same from the road. I am planning on doing three colors with five sections per color so going up and down won't work. Link to comment Share on other sites More sharing options...
-klb- Posted December 20, 2013 Share Posted December 20, 2013 But pixels would. Link to comment Share on other sites More sharing options...
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