Foundry 49: Chvorinov's Rule to Avoid Shrinkage Defects in Simple Castings
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- Опубликовано: 7 сен 2024
- Chvorinov's Rule is used to aid in calculating solidification time for castings. It can also be used as an aid to design feeders to solidify after the casting so the feeder can supply metal to the cooling casting.
This is a very light review of Chorinov's Rule and how it can be used by backyard casters to avoid shrinkage.
Chorinov established a modulus of volume of a casting or casting part divided by the area which cools that casting. Logic tells us that for the same volume, less cooling area results in a slower cooling time. So if you know the modulus of your casting, you can avoid shrinkage by making the gate and a feeder to have a higher modulus. Then the gate will cool slower than the casting so the last shrinkage of the casting will be fed by the still molten gate and avoid excess shrinkage in the casting. There is always shrinkage as a hot part cools. By the same token, a feeder supplying the gate should have an even larger modulus so it will freeze last and further protect the casting against a collapsed gate. A bob on the far end of a casting from a feeder can also protect from shrinkage but it is at a disadvantage because it will start cooler due to etal cooling as it fills the casting.
Summary: the part with the smallest modulus will freeze first. If a casting is supplied through a feeder designed to freeze last (has the larger modulus) then the feeder will start out hotter and will protect the casting.
A few key Modulus equations:
Cube - a/6
Sphere - d/6
Long Bar - a*b/(2*(a+b)
Rectangular bar - a*b*c/(2*(a*b+b*c+a*c))
Cylinder - r*h/(2*(r+h))
Pipe - t*L/(2*(t+L))
Any similarities?
A very good Sulzer paper from 1966
massalotando.f...
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I found this video very informative. I am just beginning to do some casting. I had no idea you needed to do a bit of design on the feeder portion . Glad to have found this before I learned bad habits. Thanks for the time you took to do this
Cheers from Canada
Thanks for the kind comments.
You can do a lot of great casting ignoring this rule. I usually ignore it. but when you have a shrinkage problem this will generally solve it. Hopefully no one avoids casting thinking they have to do this kind of work first. However once you do it a few times it really is easy. You can be sloppy and get better results than not trying at all.
Cast away...
👍👍😊👍👍 excellent summary. Very helpful explanation. Thank you 🙏....... Joel. (I saved this one in my playlist for future reference 😊)
Thanks for the good words. After a little practice it is surprising how easy it is to estimate a feed system size which will eliminate shrink defects. If you're casting at home a little extra metal in the feed system is not costly and can be remelted.
it would seem to me that we can leverage the power 3D cad* to subdivide the pattern appropriately and calculate the Modulus for each part. For example SolidWorks (that's what I have) is very accurate in calculating volume and surface area for any given object. You, of course, can put all the risers runners etc into your 3D model of the pattern and do exactly what is shown in the video fairly quickly and accurately. I really must try it!
At the same time thank you Andrew for this straight forward explanation of Chvorinov's Rule, of which up until this time, I never heard of.
* (for those of us that have 3D CAD --- and there are fairly inexpensive programs out there for those who don't have 3D cad yet; a _very powerful addition_ to the arsenal of homeshop machinists and backyard foundrymen)
Absolutely 3D CAD can be used to calculate the modulus. Of course it's the exposed surface area and not just the area of a particular component.
However after you have used the rule a few times you see extreme accuracy is not very important. Making a feeder larger is generally very easy and economical in home work.
Also, in my foundry I often am casting off an original part which I don't have modeled, and usually does not lend itself well to modeling.
Thanks for commenting!
Interesting stuff, Andy! You make the math look simple; and the QST emblem turned out great! (Similar on GJ)
Thanks for the comments! The math really is simple if you keep the shapes simple. And although you lose a little accuracy simplifying the shapes you gain a lot in avoiding errors. Like in high school when doing an algebra problem and Train A must be going 527.5 mph. If you simplified things and got 50 mph you are less accurate but much closer to the real answer.
Thanks for this, will help when I go to do my big casting.
It could help, I hope so. I find it helps with small castings.
Interesting stuff, thanks
Thanks, Jeff. Sorry it got so long. I don't think I have a future in public speaking.
An interesting thing, if you make patterns for risers, gates, etc. once you've calculated the modulus it does not change so I write it on the pattern piece.
@@andrewmartin4258 wouldn't pattern draft, radiused edges, etc. affect the modulus? Or maybe it all kind of cancels out in the end since all the other pattern bits would have draft too...
@@tobhomott It's an interesting subject. When you see that the modulus of a 2" square cube and 2" diameter sphere and 2" diameter cylinder 2" long are all 1/3" it seems rounded corners don't have much effect on the cooling time.
I don't look at the modulus as an exact number, and just using approximate numbers makes it doable, and gets you close. At least for what I do I can estimate a complex shape then make sure the feeder modulus is 20% or more larger than the pattern modulus, or 50% more, and feel comfortable the pattern won't shrink.
This little piece had me worried and my gate only matched the modulus of the center disk. I ignored the wings which would remove area and make the disk modulus larger. I remeasured my gate and did a more accurate modulus calculation and got 0.235". Even removing the wing attachments on the disk I think the gate is still larger.
I'm getting a mold ready to pour the same thing with a smaller gate to see the shrinkage, if any.
@@tobhomott the Chvorinov's Rule is a ratio of volume over the surface area. So yes radius, curves etc. Which change the surface area change the ratio.
@@askquestionstrythings Sorry I didn't see this response earlier.
I think one must be careful in trying to apply the modulus too strictly. It is generally accepted that a cube, sphere, and "square" cylinder all have the same modulus, "D" divided by 6. The outside corners of the cube do not add appreciably to cooling and the centers freeze at about the same time.
This is a marvelous thing for backyard casters. It lets us estimate modulus of simplified shapes and ignore little indents and protrusions. A ball peen hammer head, for instance, can be estimated as a cylinder a hair smaller that the large end, or the large end only can be taken as a simple cylinder even though it may be tapered. This gets one close to a modulus to be used to calculate an appropriate gate and feeder. Modulus increasing from mold to gate and gate to feeder.
It may sound silly but I wanna know what did you do with modulus you calculated?
It's not silly. No reason to calculate it if you can't use it.
The modulus is indicative of the time to cool to solidification. Therefore the modulus of the feeder system should be larger than the modulus of the part being cast. Then the part being cast can draw liquid metal from the feeder as the part cools and limit shrinkage in the cast part.
If you look at this video ruclips.net/video/BS_Zgdvny0g/видео.html , it shows the shrinkage coming out of a blind riser upstream of the gate.
@@andrewmartin4258 wow, thanks Andrew, now it's crystal clear..
Hi Andy. Don't you love it when people ask you questions about something you said over a year ago? 😄
I'm curious, at 4:53. Why is the area of the top of a cylinder not pi-r-squared? You know my math skills, be gentle 😄
On the contrary, I love it when people ask questions. For every person like you who asks a question there are many who did not ask that question, some being afraid of appearing dumb. So the question does not get asked. Out of 2,000 views of this video there may be 100 or more guys who had the same question. Thanks for asking.
The area is indeed pi-r-squared. The A used in the video is the diameter, which is twice the radius. I was trying to demonstrate that the modulus of the cylinder is equal to that of the cube so I needed to use the diameter. The diameter is two times the radius. The radius is from the center of the circle to the circle. So since the diameter is twice the radius, the diameter squared is four times the radius. That's why the area equation for a circle is pi-r-squared or pi-d-squared/4.
@@andrewmartin4258 Never let it be said that I'm afraid to look dumb :-D Thanks
@@swdweeb Let me affirm you have demonstrated you are indeed fearless!
@@andrewmartin4258 hey wait... should I be offended at that? 😃
@@swdweeb Oh, I don't think so...
Having friends like you makes me feel better about myself.😎
Maybe I missed it somewhere but what is the purpose of the module? Do you want all parts to have the same module? Is that the purpose?
Apparently I was not clear. I edited the text under the video to attempt to add clarity.
Basically the larger the modulus the slower a part cools. You want your casting to freeze first and the feed system to supply makeup metal as the casting cools and shrinks. So the modulus of the gate, and then the modulus of a feeder should each in turn be larger. Then the casting should freeze first, then the gate, then the feeder.
That was a big omission in the video. Thanks for bringing that to my attention!
@@andrewmartin4258 thank you for your clarification and thank you for the hard work put into the video. Much appreciated.
So the blind riser modulus should be equal to the sum of the moduli of the parts being fed?
Does that include the gate?
I think I'll have to curse my way through that Chcovinovs paper.
BTW, what was going on with that bar that you cast first, the circle on the surface, I mean.
260 pages!??!
Never get your education from a You Tube poster. Ha Ha Ha.
I think the modulus of the blind riser only need be larger than the modulus of the part it is feeding, and so on cascading down. The idea is progressive solidification, so the last parts should have the smallest modulus. They should freeze first. I did not calculate the wings because experience told me they would not shrink, it would be the center. If the gate can feed the center, the center won't shrink. If the riser can feed the gate, the gate won't shrink. My runner was in trouble.
Mr. Chvorinov was way ahead of his time. I admire his genius. The more you read the more you learn.
That was an ingot cast only to get rid of some metal in the crucible to give me a better pouring angle. It's a new ingot mold and has some rust in it. That was the rust boiling and contaminating the ingot.
@@stoneomountain2390 Ten pages and you're about done. It is good reading, you can grasp the concept of what the modulus is and that opened my eyes as to how to reduce or eliminate shrinkage.
@@andrewmartin4258
Thanks for the info, and the reassurance.
I hope you enjoy your weekend.
Sorry about your friend 1/2 cup. Thank you for info.
Thank you. Though I never met him, we had become good friends.
for the center part you really need to include all the parts.
Note I'm ignoring the letters (which lower the volume and increase the surface area) you basically have two cylinders stacked.
the height was relatively easy to read off the video, h1 = 10/32in h2 = 6/32 in h_total verifies h1+h2 = 0.5 in
the diameter of the bottom part I had to estimate as being 2/8in larger that the diameter of the top part. or 2 5/8 in (2.625in)
volume is v1+v2
area is a little complicated as you subtract to overlap area
after simplifying I get a Vt/At = (d1^2 *h1 + d2^2 *h2)/(4 *d1 *h1 + 2 *d2^2 + 4 *d2 *h2) (
Thanks for taking the time! After you got significantly different results I felt the need to go back and check my work.
First I want to reiterate that my goal is to keep the method as simple as possible primarily so I will use it. If it becomes overly complicated then I won't use it and it is of no value. Secondly other backyard casters may find it is useful if it is easy enough to understand and use.
I have to respectfully disagree on the assumption that all minor surface shapes are of importance in calculating the modulus. The lettering, for instance, and any inside corners, although they add area they do not add to cooling. The sand does not flow so there is only conduction and inside corners soon get hot and do not add to cooling. Chvorniv and later researchers have shown that the modulus of a square box, and inscribed cylinder, and an inscribed ball all have the same modulus. The exterior corners also add little to cooling, they quickly cool on a square box and the liquid remaining in the middle is a ball.
Using your numbers for size, I calculate the volume is 2.27 cu in. The top and bottom add to 10.82 sq in, and the rims add to 3.67 sq in for a total area of 14.49 sq in giving a modulus of 0.157 in. It would be great if you could check my work to determine where we disagree.
Actual measurements are a little different. Top is 2.35 in and bottom is 2.46 in. I estimated 2.375 in to be about 1/3 of the difference. With actual measurements I get a modulus of 0.162 in.
Now, if I deduct the area of the wing attachments, I get a modulus of 0.171 in. If I also deduct the area of the gate contact I get a modulus of 0.180 in. In practice I don't want to go to that level of detail or I won't do it.
I feel confident that Chvorinov's Rule can be applied loosely and simply and still give good results.
A little fudging and a little judgement go a long way. For instance, I neglected the entire top area of my gate because it is mostly in contact with the mold or feeder and the small part left is sandwiched between the feeder and the mold. That common inside corner area does not contribute to cooling much to any of the three bodies. It helps account for neglecting the wing attachments.
I await your review of my areas and volume.
@@andrewmartin4258 surface area does matter. consider the limit as A approaches infinity of V/A is zero. Thus we see that as the surface area approaches a very large number the modulus will decrease. but note: I ignored it in making calculations here as I don't have a way to get that number (you could possibly pull the volume and surface area from the CAD program the part was modeled in).
I'll come back when I have a moment for the rest of the calculations. it's minutes o'clock and time for me to make some things happen.
@@andrewmartin4258 the total volume I calculate from my estimated sizes for the center is 2.399 in^3
using your more precise numbers rather than my estimates
d1 = 2.35 in, d2 = 2.46, h1 = 0.3125 in , h2 = 0.1875 in
I get a total volume of 2.247 in^3
the volumes of the two cylinders add V1+V2.
Vcyl = π*r^2*h and substituting d = 2*r -> r = d/2
Vcyl = π*(d/2)^2*h -> Vcyl = π/4*d^2*h
thus the total volume of two stacked cylinders of different sizes is
Vtotal = π/4*d1^2*h1 + π/4*d2^2*h2
the surface area for stacked cylinders is slightly more complicated.
you have the area of the circle for the top of cylinder 1 plus the area of the rectangle making the side of cylinder 1
then you add the area of the circle for the bottom of cylinder 2 plus the area of the rectangle making the side of cylinder 2
now we have a little area to add where the two meet and cylinder 2 is larger than cylinder 1 (a doughnut)
we take the area of the circle on the top circle for cylinder 2 and subtract the area of the circle on the bottom of cylinder 1 (the overlap)
(at this point I just noticed I forgot to add the area of the two rectangles which threw off my simplification)
Ac = circle 1 + circle 4 + [circle 3 - circle 2] + rectangle 1 + rectangle 2
Ac = (π/4*d1^2) + (π/4*d2^2) + [(π/4*d2^2) - (π/4*d1^2)] + π*d1*h1 + π*d2*h2
combining and simplifying V/A (center without the wings)
Vt/At = (d1^2 *h1 + d2^2 *h2)/(4 *d1 *h1 + 2 *d2^2 + 4 *d2 *h2)
I get 0.169 with the actual measurements and a minimum required riser/feeder modulus of 0.185 (20% larger)
0.2*(Vpart/Apart)^2 +(Vpart/Apart)^2 = (Vriser/Ariser)^2
I did have an error in my first run through; a bookkeeping error on my part. I would say with my revised calculation and your revised numbers we are likely within a margin of error now 0.169 vs 0.171 .
If you using Fusion 360 for modeling you can get the physical properties from the Properties palette. right click the model and select properties from the context menu. You should be able to quickly read off the Surface Area and Volume. Most cad programs have a similar option. Much faster less error prone and more accurate than cobbling together basic shapes to approximate the Surface Area and Volume.
@@askquestionstrythings Glad you found your error. But we're not having an arithmetic contest. I failed to give you the true thicknesses which are 8 mm for the top and 4 mm for the bottom. That should make our numbers come out.
In calculating areas I only calculated the area of the large circle and doubled it. That exactly accounts for the area of the small circle plus the donut. I didn't care what their individual values were.
Most of my patterns are done in wood, or off of original cast parts. So I need a simple way to calculate a modulus which will help me size a feeder.
Our primary area of difference is in the relevance of surface variations which I consider to be insignificant and you don't. Things like the lettering, and even the small step between the small circle and the large one. If you consider an isotherm surface slightly away from the body in question it is a smoothed version of the body. The farther you go away the more spherical the isotherms become. A groove soon becomes lost in the isotherms. And it is the isotherms which show how the energy is leaving the body, and the actual rate of energy loss determines the time to freeze.
To avoid error I believe it is important to not get lost in the details. If we have a shaft with Acme threads I am going to estimate the shaft diameter to be the O.D. less maybe 1/3 of the thread depth, each side, so a 2" shaft with 0.09" deep threads I would guess to be 1-15/16" OD and never look back. The sand 1/4" away from the shaft can't tell there are threads. If one calculated the sides of the threads one might be tempted to double the shaft surface area (depending on the pitch). That would be accurate for painting but would be an error to calculate a realistic modulus, in my opinion.
I'm currently preparing to copy a drill press table for a blacksmith hand crank drill press. I don't want to model it. And I don't even think I should remove the through slots in calculating a modulus. Of course whatever I calculate I'll be generous in adding modulus for the feed system.
Thanks for your continued interest.
@@andrewmartin4258 I don't think communicated well what I was saying about the Surface area mattering. From the limit as the surface area increases the modulus will decrease. Therefore by simplifying the surface area you *overestimate* the part modulus (i.e. the actual modulus is smaller than the estimate) which means you will have a larger riser/feeder than you would have if you included all of the surface areas. As long as you have an excess of metal to fill to the total volume of the casting, runners, riser/feeder this is a good situation. For backyard and small shops simplifying the surface area is typically a good thing (unless you cant fill the system and run out of metal).
On the other hand, *you don't want to underestimate the volume* as a reduced volume has a much bigger effect on the modulus. Underestimating the volume results in underestimating the modulus and results in a riser/feeder smaller than you need.
For the shaft with ACME threads, I would argue that it's better to estimate Chvorinov's rule for the shaft as just a cylinder the size of OD. This would overestimate the volume and underestimate the surface area resulting in a larger modulus and a larger feeder/riser.
In the case of a non-CAD-modeled part, I agree that cobbling together an estimate based on the sums of simple shapes is best. For the drill press table estimating based on a series of stacked rectangles (based on the thickest cross sections) and not including the slots in the Chvorinov's rule calculation should result in overestimating the volume and underestimating the surface area resulting in a larger modulus and a larger feeder/riser... i.e. a better chance of feeding the table. (still requires a 20%-25% increase in the feeder/riser e.g. 0.2*(Vpart/Apart)^2 +(Vpart/Apart)^2 = (Vriser/Ariser)^2)
Martin how can a get in contact with you? I need to show you my casting project and the surface shrinkage problem I am getting on my aluminum flag pole top. 12 inch long part. Basically a sphere with a long tapered cone on it. The surface shrinkage always occurs on the sphere section on the top section of the part. I would like to share some pictures of my problem and a workout a solution. Are you on Face Book?
Lots of helpful guys on The Home Foundry forum. I post there as well.
@@andrewmartin4258 Is that a fb group? Can you send me a link please?
@@gulf4yankee forums.thehomefoundry.org/index.php