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Seeking a specific formula to allow for materials shrinkage

XNotLost

New member
Apr 6, 2019
3
First, let me apologise if this is in the wrong thread, wrong place, or wrong forum.

I've not had any good experiences so far getting an answer I can understand, because my maths education didn’t really prepare me to even begin to try to figure this one out!

I'll try to explain the situation as best as I can.

I have a material to be used in molding objects, which shrinks approximately 15% as it dries.

I have objects that are to be molded in this material, and they must end up at a specific size. These objects can be created at any size necessary. They are to be 3D printed, and will be used as the forms around which the thermoforming process will create the mold.

One of the properties of the molding material is that it shrinks about 15% as it dries.

I will be making a set of vacuum thermoformed molds, and my original objects will need to be sized in such a way as to create a thermoformed mold that is approximately 12%-15% larger than the final object, to allow for shrinkage.

For example, If I make the mold at 2” wide to start, the wet material will go into the mold, and then will come out of the mold and dry, and then end up at 1.85” wide. I need to make the mold slightly bigger than 2” for it to come out and shrink to dry at 2”. If I have a part that needs to end up at 6.75” wide, then I need to make it slightly bigger than 6.75” wide for it to come out at the correct width. I just don't know how to figure out how to do this.

I would like to find a maths formula that allows me to enter a value for the finished size of an object, allowing for this 15% shrinkage from the original condition, that will give me the starting size for the mold, no matter what my finished size is.

One of the things that makes this much easier for me, is the means by which my form will be printed. I can enter one dimension and the others will automatically be generated, proportionally. So, if I know I need a mold-form to end up at say, 1.75" wide, I can enter that dimension, and the length and depth will be automatically generated.


I have to say, this isn’t the kind of maths I am familiar with.

For some of you, this will be the easiest thing in the world. For me, not so much. All my maths teachers wrote a problem on the board, and then the answer, without ever putting anything that goes in between on the board. When asked what the steps were, I was always told "you should know how to do this already" I got suspended from school once for telling this one teacher that it was his job to teach us the stuff in the middle, not just how to write a problem and answer, and expect us to "already know" how to go from one to the other. No wonder so many people don't like maths!
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
617
North Texas
A 2” form shrunk by 15% will end up at 1.7”, not 1.85” ... that is, the end product will be 85% of the original form’s size.

0.85(2) = 1.7

if you want to have an object end up at size $y$ starting out at size $x$ given a 15% shrinkage, then ...

$0.85(x) = y \implies x = \dfrac{y}{0.85}$

so, if you want to end up with a 2” wide object, you need to start with a form of size $\dfrac{2”}{0.85} \approx 2.35”$

note that the original form in the above case is a bit over 17.5% larger than the end product.

to end up with an end product of 6.75” with a 15% shrinkage, you start with a form of size $\dfrac{6.75”}{.85} \approx 7.94”$, again a little over 17.5% larger.

A final note, all of this assumes the shrinkage is constant in all directions and thicknesses for a 3D object. You may have to make a few experimental runs to ensure you end up with the size desired.
 

XNotLost

New member
Apr 6, 2019
3
A 2” form shrunk by 15% will end up at 1.7”, not 1.85” ... that is, the end product will be 85% of the original form’s size.

0.85(2) = 1.7

if you want to have an object end up at size $y$ starting out at size $x$ given a 15% shrinkage, then ...

$0.85(x) = y \implies x = \dfrac{y}{0.85}$

so, if you want to end up with a 2” wide object, you need to start with a form of size $\dfrac{2”}{0.85} \approx 2.35”$

note that the original form in the above case is a bit over 17.5% larger than the end product.

to end up with an end product of 6.75” with a 15% shrinkage, you start with a form of size $\dfrac{6.75”}{.85} \approx 7.94”$, again a little over 17.5% larger.

A final note, all of this assumes the shrinkage is constant in all directions and thicknesses for a 3D object. You may have to make a few experimental runs to ensure you end up with the size desired.
Thanks, but let me make something clear first. I made up all numbers, and just guessed at what might be what.

I almost understand your reply, but I don't know where you get the 17.5% from. I didn't mention anything remotely like that.

Now, it may just be the way computers and forums write things, but all that

" $ \ d frac "

mess means less than nothing to me. I don't know what a dollar sign means other than money. All of the other symbols and notations you used are not helpful and only confuse me. Is there any way that you could just say it like this:

In order to find how much bigger your form should be to get your final product the right size, multiply by <whatever>

Pretty please?

And the reason for me asking this is, I am exactly trying to avoid having to make any extra runs of any parts to create the molds. Experimentation is not in the cards. I am trying to plan ahead, because extra 3D prints cost money that is budgeted elsewhere.
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
617
North Texas
The dfrac is just code to put the numbers in a math friendly fraction form in the actual post.

Once again, if you desire to end up with a 2” size for your end product given a 15% shrinkage, then the 2” will be 85% of what you need to start with ...

85% of the initial form size = 2”

So, initial form size = 2”/.85 , which is about 2.35”

don’t worry about the 17% if you don’t get where it came from.
 

XNotLost

New member
Apr 6, 2019
3
The dfrac is just code to put the numbers in a math friendly fraction form in the actual post.

Once again, if you desire to end up with a 2” size for your end product given a 15% shrinkage, then the 2” will be 85% of what you need to start with ...

85% of the initial form size = 2”

So, initial form size = 2”/.85 , which is about 2.35”

don’t worry about the 17% if you don’t get where it came from.
Okay, just to confirm that I get what you're telling me...

If I have to have the end result of the dried molded part at 3" high (let's just assume that all the other dimensions will shrink at the same rate, which seems to be what this stuff does) then I have to take that 3" final part, and divide it by .85 to get the size of the part I have to print to make the mold.

So, 3" divided by .85 equals 3.529 or 3.53"

Do I have this right? That "divided by .85" works in every case? Seems like a half inch out of three inches is a lot more than 15%

Much appreciated!!