Weight for the BOM

[cz2062]

Hi,
I am doing our team's bill of materials. I was specifically appointed to doing the cost part and then I was going to integrate it with the weight spreadsheet, but then I realized through the chaos of the build season, we missed several weights. I was wondering if I could collaborate with some of you who have the same items. For one I need the weight of a piece of item. From there I can figure a unit weight and apply it to our pieces.

Thanks for the help...

[Mike Martus]

Weight is not required for the BOM. Weight IS required for the inspection.

You need cost correctly calculated - see FIRST Rules as they apply to smallest quantity punchable. In some cases if the item is only available in say a 4 ft x 8ft sheet you MUST account for the entire sheet even you only used 1 inch. This is a consideration when building the robot and selecting materials.

Also note any Speed controllers above the kit amount must also be calculated in the total cost of the robot.

Read the rules.

Inspectors like it when you provide a NEAT excel spreadsheet, as it shows accuracy and thought. What is really a mess is a BOM written on a scrap piece of paper or the team that has a BOM but it is electronic and they have no printer.

[Richard Wallace]

Mike is right, you don't need to show weights for each line-item on your bill of material.

Please read all of section 8.3.3 and especially the examples given in section 8.3.3.1 of the 2008 FRC Manual before you complete your bill of material spreadsheet. After you've studied that section, please post specific questions here if you need more help.

I'll be at the St. Louis Regional and will be glad to answer questions about your bill of materials if you'll ask for me at pit administration early on Thursday. Well done bills of material make robot inspection go much faster.

[cz2062]

I did read the rules, but when I saw the BOM template provided at FIRST, it included weight as a column. Thanks for your replies...

So I take it you guys wouldn't have the weight of a piece of item then?

[Mike Martus]

How much any part weighs is not important once the total weight of your robot meets max weight.

The time for weights on specific parts and materials was in the planning and design stages.

The real issue is when the robot exceeds the weight limit.

REMEMBER! Drilling holes in Alum. is not a weight savings of any consequence.

[dlavery]

Originally Posted by Mike Martus

REMEMBER! Drilling holes in Alum. is not a weight savings of any consequence.

Oh, I beg to differ. Approximately 100 one-half-inch holes in 1/8" aluminum will remove a pound from your robot. There are many times when that pound can mean the difference between passing or failing the weight test. And 100 holes can be strategically spread around on almost any robot without too much effort.

We have a power drill with a 1/2-inch bit permanently friction-welded into the chuck (affectionately known as "halfie") that we keep handy at all times, just in case.

Of course, we may have taken that philosophy to an extreme the year that we put 1,248 holes in our frame (we had to count them all. and now we know how many holes it takes to fill the Albert Hall).

-dave


.

[Richard Wallace]

Originally Posted by dlavery

...
Of course, we may have taken that philosophy to an extreme the year that we put 1,248 holes in our frame (we had to count them all. and now we know how many holes it takes to fill the Albert Hall).

I'll bet it took longer to de-burr all of those holes than it took to drill them.

[lukevanoort]

Originally Posted by dlavery

Oh, I beg to differ. Approximately 100 one-half-inch holes in 1/8" aluminum will remove a pound from your robot.-dave


.

I think your math may be a bit off. Since a half inch hole removes pi * .25"^2 * .125" or .0245 cu in, and aluminum has a density of .0975 lbs/cu in, each hole removes .002393 pounds of material. 100 is only .2393lbs. 100 1" holes does lose 1lb though. A note to those needing to lose weight, 100 1" holes take a while to make; slicing and dicing any steel you may have will be much more time-effective.

[Mike Martus]

Dave.... you are correct.... however there are usually other items that weigh more and if removed will not weaken the structure.

Consider.......
Using Alum washers in place of steel
Using Alum clamp collars or even elim. the collar and replacing with a roll pin
Best of all.... using alum rivit in place of steel fasteners

I also like letting the air out of the IFI Traction wheels.....

Other weight saving ideas??????

[Ian Curtis]

Originally Posted by lukevanoort

I think your math may be a bit off. Since a half inch hole removes pi * .25"^2 * .125" or .0245 cu in, and aluminum has a density of .0975 lbs/cu in, each hole removes .002393 pounds of material. 100 is only .2393lbs. 100 1" holes does lose 1lb though. A note to those needing to lose weight, 100 1" holes take a while to make; slicing and dicing any steel you may have will be much more time-effective.

[WRONG]Your math is right, but a quarter inch is not a half inch. [/WRONG]

However, Luke is certainly right in terms of time. On Thursday you're far better off removing non vital assemblies and parts than cutting holes if you're a pound or more over.

[lukevanoort]

Originally Posted by iCurtis

Your math is right, but a quarter inch is not a half inch.

However, Luke is certainly right in terms of time. On Thursday you're far better off removing non vital assemblies and parts than cutting holes if you're a pound or more over.

Indeed, a quarter inch is not a half inch, but the equation for the area of a circle is pi*radius^2, and drill bits are measured in terms of diameter. So, a half inch bit has a quarter inch radius, and thus the area of the hole is pi*.25^2.

[Richard Wallace]

Originally Posted by iCurtis

Your math is right, but a quarter inch is not a half inch. ...

Originally Posted by lukevanoort

Indeed, a quarter inch is not a half inch, but the equation for the area of a circle is pi*radius^2, and drill bits are measured in terms of diameter. So, a half inch bit has a quarter inch radius, and thus the area of the hole is pi*.25^2.

Most of us learn the pi-R-squared formula for circular area in high school.

Engineers prefer the formula Area = (pi/4) * diameter^2, because measurement of a circle's diameter is generally more accurate than measurement of its radius. The diameter measurement only requires finding two (diametrically) opposite points on the circle, while a radius measurement requires finding the circle's center.

The pi-R-squared formula always reminds me of the old joke about the college kid who comes back to the farm and is asked to say something smart. When he replies "pi-R-squared", his father the farmer says, "son, you're still as dumb as the day you left here. Everyone knows that pie are round -- cornbread are squared."

[Ian Curtis]

Originally Posted by lukevanoort

Indeed, a quarter inch is not a half inch, but the equation for the area of a circle is pi*radius^2, and drill bits are measured in terms of diameter. So, a half inch bit has a quarter inch radius, and thus the area of the hole is pi*.25^2.

Heh. Yep. You're right. I think I should probably put down me calc book and go find a nice grade school geometry book.