Math question for the experts.
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To calculate volume, wouldn't you need to know how deep the holes are? The thickness of the material the holes are drilled into? Without that, all you could calculate is the surface area of the holes.
Originally Posted By: Hirev
What would be the total volume in diameter? Does that make since?
No, it does not make sense. Volume is something that can be calculated for a 3-dimensional object such as a block, cylinder, or a sphere. A hole (circle) is 2-dimensional, and therefore does not have any volume. A 2-dimensional object has an area though.
The area of a 0.1750 diameter hole (circle) is about 0.024 square units, so eight such holes would have a combined area of 0.192 square units.
Now, if you know how deep the hole is, then you can calculate the volume (in cubic units).
What would be the total volume in diameter? Does that make since?
No, it does not make sense. Volume is something that can be calculated for a 3-dimensional object such as a block, cylinder, or a sphere. A hole (circle) is 2-dimensional, and therefore does not have any volume. A 2-dimensional object has an area though.
The area of a 0.1750 diameter hole (circle) is about 0.024 square units, so eight such holes would have a combined area of 0.192 square units.
Now, if you know how deep the hole is, then you can calculate the volume (in cubic units).
Originally Posted By: Hirev
Let say you have eight .1750 diameter holes. What would be the total volume in diameter of all eight combined? Does that make since?
"Total volume in diameter" doesn't make sense. Are you looking for volume or diameter (or something else)?
Let say you have eight .1750 diameter holes. What would be the total volume in diameter of all eight combined? Does that make since?
"Total volume in diameter" doesn't make sense. Are you looking for volume or diameter (or something else)?
Originally Posted By: Hirev
Ok, perhaps total diameter of all eight holes?
Just multiply by 8.
But maybe you should first explain what you're trying to do because it's a bit of an odd question, still. I have a feeling you're really looking for something else entirely.
Ok, perhaps total diameter of all eight holes?
Just multiply by 8.
But maybe you should first explain what you're trying to do because it's a bit of an odd question, still. I have a feeling you're really looking for something else entirely.
This makes me think of oil filter holes (like Fram).
Are you looking for the volume they can pass?
Are you looking for the volume they can pass?
You want the total area.
pi(r)^2 or 3.14(radius)^squared
radius is 1/2 the diameter.
pi(.1750/2)^2 = .0240528
8 times that is 0.192423
http://www.wolframalpha.com/input/?i=8%28pi%28.1750%2F2%29^2%29
To find the equivalent hole size take 2 x the square root of (0.192432/pi)
http://www.wolframalpha.com/input/?i=2%28sqrt%28.19243%2Fpi%29%29
Answer is .494 or just under 1/2
pi(r)^2 or 3.14(radius)^squared
radius is 1/2 the diameter.
pi(.1750/2)^2 = .0240528
8 times that is 0.192423
http://www.wolframalpha.com/input/?i=8%28pi%28.1750%2F2%29^2%29
To find the equivalent hole size take 2 x the square root of (0.192432/pi)
http://www.wolframalpha.com/input/?i=2%28sqrt%28.19243%2Fpi%29%29
Answer is .494 or just under 1/2
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Originally Posted By: NHGUY
3.534 square inches?
NO, and the OP didn't specify units (inches, mm, cm, feet, etc)
3.534 square inches?
NO, and the OP didn't specify units (inches, mm, cm, feet, etc)
I'm looking at a Wix oil filter 57356 for the new suggested filter per wix site. It has Eight .175 diameter holes in the base plate.
I'm also looking at their old suggested part number for the same application a Wix 51356. It has Eight .245 diameter holes in the base plate. How to the figure eight .175 diameter holes are now enough?
I'm also looking at their old suggested part number for the same application a Wix 51356. It has Eight .245 diameter holes in the base plate. How to the figure eight .175 diameter holes are now enough?
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Originally Posted By: Loobed
You want the total area.
pi(r)^2 or 3.14(radius)^squared
radius is 1/2 the diameter.
pi(.1750/2)^2 = .0240528
8 times that is 0.192423
http://www.wolframalpha.com/input/?i=8%28pi%28.1750%2F2%29^2%29
To find the equivalent hole size take 2 x the square root of (0.192432/pi)
http://www.wolframalpha.com/input/?i=2%28sqrt%28.19243%2Fpi%29%29
Answer is .494 or just under 1/2
Yes that is what I'm after thanks all!
You want the total area.
pi(r)^2 or 3.14(radius)^squared
radius is 1/2 the diameter.
pi(.1750/2)^2 = .0240528
8 times that is 0.192423
http://www.wolframalpha.com/input/?i=8%28pi%28.1750%2F2%29^2%29
To find the equivalent hole size take 2 x the square root of (0.192432/pi)
http://www.wolframalpha.com/input/?i=2%28sqrt%28.19243%2Fpi%29%29
Answer is .494 or just under 1/2
Yes that is what I'm after thanks all!
I think your problem can't be just solved with total area of 2 different scenarios.
1 equipment with 10 holes with total 10 sq inches will flow less with more pressure than the equipment with 1 hole with 10 sq inches. There is a thing called boundary layer to the flow and the flow will be faster in the center of the hole than around the circumference of the hole. etc
1 equipment with 10 holes with total 10 sq inches will flow less with more pressure than the equipment with 1 hole with 10 sq inches. There is a thing called boundary layer to the flow and the flow will be faster in the center of the hole than around the circumference of the hole. etc
I use that filter - the 57356 is the correct filter for honda applications. The 51356 overburdened my 1.5 L15 GE-08 fit engine with variable volume/efficiency oil pump. I dont know why. We can assume the multiple small inlet holes ( ala Purolator) are to shear thick, cold oil to preserve the filter media pleating and aid cold flow and warmup, and prevent media hydro- shock (at least until a wix engineer or honda engineer explains differently). The "nice looking" to a novice and layman 51356 had far beyond average, oversize inlet holes most like ly designed for small engine low pressure oil pumps on its application list. What do you think?
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Originally Posted By: Hirev
I'm also looking at their old suggested part number for the same application a Wix 51356. It has Eight .245 diameter holes in the base plate. How to the figure eight .175 diameter holes are now enough?
.175 / .245 = 0.71
The one with smaller holes has only %71 the open area of the one with the larger holes. But considering the biggest restriction is the filter itself and the pressure of the bypass valve, these numbers aren't everything.
I'm also looking at their old suggested part number for the same application a Wix 51356. It has Eight .245 diameter holes in the base plate. How to the figure eight .175 diameter holes are now enough?
.175 / .245 = 0.71
The one with smaller holes has only %71 the open area of the one with the larger holes. But considering the biggest restriction is the filter itself and the pressure of the bypass valve, these numbers aren't everything.
Originally Posted By: Hirev
Let say you have eight .1750 diameter holes. What would be the total volume in diameter of all eight combined? Does that make since?
Infinite
Let say you have eight .1750 diameter holes. What would be the total volume in diameter of all eight combined? Does that make since?
Infinite
Do you mean "area" of the holes combine?
The 8 x PI x radius (0.175) square will be the answer.
But if you are looking for flow volume. The amount of fluid that can flow through 8 0.175" holes is different than one single hole of the same volume. Fluid dymanic is not linear. Someone with mechanical engineering background should be able to tell you more about it. I have a feeling that it has to do with viscosity of the said fluid too.
The 8 x PI x radius (0.175) square will be the answer.
But if you are looking for flow volume. The amount of fluid that can flow through 8 0.175" holes is different than one single hole of the same volume. Fluid dymanic is not linear. Someone with mechanical engineering background should be able to tell you more about it. I have a feeling that it has to do with viscosity of the said fluid too.
Originally Posted By: Hirev
I'm looking at a Wix oil filter 57356 for the new suggested filter per wix site. It has Eight .175 diameter holes in the base plate.
I'm also looking at their old suggested part number for the same application a Wix 51356. It has Eight .245 diameter holes in the base plate. How to the figure eight .175 diameter holes are now enough?
Both filters have the same max flow rate, according to Wix. So I'm assuming the size of these holes isn't the flow limiting factor, if that's what you're wondering. The filtering medium itself is probably the limiting factor.
I'm looking at a Wix oil filter 57356 for the new suggested filter per wix site. It has Eight .175 diameter holes in the base plate.
I'm also looking at their old suggested part number for the same application a Wix 51356. It has Eight .245 diameter holes in the base plate. How to the figure eight .175 diameter holes are now enough?
Both filters have the same max flow rate, according to Wix. So I'm assuming the size of these holes isn't the flow limiting factor, if that's what you're wondering. The filtering medium itself is probably the limiting factor.
At one time, the average High School graduate would have been able to answer the question of total area. Now it's believed that such a question requires a math expert for the correct answer. Sadly, such is the state of our public educaton.
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Originally Posted By: Brian Barnhart
At one time, the average High School graduate would have been able to answer the question of total area.
But we don't know if OP is a high school graduate.
With that said, I think I learned how to calculate the area of a circle in 4th grade, but did I remember how to do it right now? Heck no. I had to google it. You don't use it, you lose it.
At one time, the average High School graduate would have been able to answer the question of total area.
But we don't know if OP is a high school graduate.
With that said, I think I learned how to calculate the area of a circle in 4th grade, but did I remember how to do it right now? Heck no. I had to google it. You don't use it, you lose it.
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