sbode
Posts: 2
Joined: Mon Apr 10, 2017 2:44 pm
Occupation: Parent

Juice Box Geometry

Hello, my son and I were conducting this experiment and we measured 7 different brands. We have completed the entire project, however, I am trying to determine how to decipher which juice box actually uses the least amount of packaging. In the variations section, it states to compare Surface Area to Volume by dividing S by V. Am I suppose to divide by the actual fluid ounces each container holds (3 hold 6 fluid ounces and 4 hold 6.75 fluid ounces?? Or do I divide by the calculated volume? Each way provides opposing answers.

For instance, I will provide the calculations for the 3 that held 6 fl. oz
box 1: S=213.64 cm2 V=187.872 cm3 so... S/V=1.1372
box 2: S=215.36 cm2 V=189.696 cm3 so... S/V=1.1353
box 3: S=204.82 cm2 V=174.276 cm3 so... S/V=1.1753

So this would mean that box 3 (when looking at the surface area and the fl. oz it holds) uses less packaging material. But if I calculate the surface area divided by the calculated volume, it seems that this box uses the most amount of packaging. Then again if I divide the surface area by the fluid ounces, it again seems as though box 3 uses least amount of material.

dcnick96
Moderator
Posts: 489
Joined: Wed Jul 25, 2007 7:59 pm

Re: Juice Box Geometry

Hi, there; and welcome to Science Buddies! This is a fun project.

For your S:V calculations, I would use the calculated volume using your measured length, width, and height (equation 2 in the procedures). I suspect the reported fluid ounces on the front of the container is less than this calculation, as they don't fill the container all the way to the top...there is a little bit of air.

Let's talk about the results of your S:V calculations. The larger the result, the larger the "S" value is, which means more packaging. Therefore, you are looking for a SMALLER number to identify which juice box uses the least amount of packaging per fluid ounce. In your results below, I would conclude box # 1 and 2 are the same, and box #3 uses more packaging.

So, why are you recording the advertised fluid ounces from the front of the container, you might ask? If I were making recommendations on setting up this experiment, I would recommend to keep this number the same (all 6 oz containers, all 10 oz containers, etc). If you mix and match the container size, you are adding an additional variable; and this might skew results. I don't know if this is true, but perhaps the manufacturer uses more / less packaging, depending on the container size.

I hope this helps. Please write back if you have further questions. Good luck!
Deana

sbode
Posts: 2
Joined: Mon Apr 10, 2017 2:44 pm
Occupation: Parent

Re: Juice Box Geometry

Thank you for your response. The issue is, that I figured the larger the S value, the more material used, however, when calculating S/V, just the opposite shows in the results. The container with the least amount of S value (when only comparing the same amount of fluid ounces) should be the one that uses least amount of material, but once I calculate the S/V it reverses that result, and makes the container with the least amount of S value the container that actually uses the most.

Again, these 3 containers hold 6 oz... (which doesn't matter I suppose)
box 1: S=213.64 cm2 V=187.872 cm3
box 2: S=215.36 cm2 V=189.696 cm3
box 3: S=204.82 cm2 V=174.276 cm3

When viewing these calculations, one would say, well box #3 uses least amount, then box #1, and box #2 uses most.
But once the S/V calculation is performed, the results are:
box 1: S/V=1.1372
box 2: S/V=1.1353
box 3: S/V=1.1753

Which then provides the answer, Box #2 uses the least amount of material, then box #1, and box #3 uses the most. So which is the correct way to decide which uses most material?

dcnick96
Moderator
Posts: 489
Joined: Wed Jul 25, 2007 7:59 pm

Re: Juice Box Geometry

Hi there. So, I should apologize, as saying "a larger S value means more packaging" isn't the full story. If the "V" value were the same across all boxes measured, this statement would be true. However, in your case, "V" for box 3 is different from the other 2. Therefore, the value of S doesn't tell the whole story, and now the S:V ratio comes into play.

By calculating the ratio, you are "normalizing" the comparison of fluid ounces to surface area (packaging used). This technique is common in this type of calculation, because it is rare for 2 objects to be exactly the same;and we must find a way to eliminate a bias that would come with those differences. Calculating a ratio does that.

So, the correct answer is to use the results of the S:V ratio (Box 3 uses more packaging) and not just the S value.

Sorry for the confusion. I hope this helps. Write back if you have more questions.
Deana