Randomness

[Distracted]

Whoa. Now my brain hurts.

[JadziaKathryn]

That's worse than trying to decipher IRS-speak.

[CX]

Just got back from a live version of the Rocky Horror Picture Show, and it was awesome. They found a very attractive actress for the part of Magenta, and we even got a short strip tease at the beginning of the show, and her breasts made another appearance toward the end of the show.

[blacknblue]

There has to be a pre-written formula for calculating the circumference, diameter, slope, and weight /resiliency ratio for those breasts.

[Distracted]

Just got back from a live version of the Rocky Horror Picture Show, and it was awesome. They found a very attractive actress for the part of Magenta, and we even got a short strip tease at the beginning of the show, and her breasts made another appearance toward the end of the show.

Thanks, CX. You made me feel much less guilty about nixing my 16 year old daughter's plans to go to a live performance of Rocky Horror at midnight last night. She went to a party instead and was home by 11.

[leslina]

Just got back from a live version of the Rocky Horror Picture Show, and it was awesome. They found a very attractive actress for the part of Magenta, and we even got a short strip tease at the beginning of the show, and her breasts made another appearance toward the end of the show.

Thanks, CX. You made me feel much less guilty about nixing my 16 year old daughter's plans to go to a live performance of Rocky Horror at midnight last night. She went to a party instead and was home by 11.

God bless! Am SO glad I'm not a mother.

[Distracted]

Heh. Some days it's not so fun. Needless to say, I'm not her favorite person right now.

[blacknblue]

"A vertical true centrifugal castig process is used to cast brass tubing that is 15 in. long and whose outside diameter = 8.0 in. If the speed of rotation during solidification is 1000 rpm, determine the inside diameters at the top and bottom of the tubing if the total weight of the final casting = 75.0 lb."

"The difference in inside radius between the top and bottom is related to the speed of rotation as follows: N = (30/Pi)[(2*g*L)/(Rt^2 - Rb^2)]^1/2
Where L = vertical length of the casting, Rt = inside radius at the top of the casting, Rb = inside radius at the bottom of the casting, g = the gravitational constant (32.2 ft/s^2), and N = the rotational speed in RPM."

Using the above equation, I determined (Rt^2 - Rb^2) = 0.007340719 ft^2

The density of brass = 0.313 lb/in^3, which I used to determine the volume of the cast part to be 239.62 in^3.

There is no illustration of the part, but none of the problems require them since the problems are all basically numbers games of manipulating the numbers you have to get the numbers you want.

[Caveat: The following all assumes that I am reading the question correctly, and that I have not gone senile. Two fairly large assumptions. ]


No doubt this is not the way they want you to derive it, but if you have the outside dimensions and the weight, plus the density, you have everything you need to just get the actual numbers. I am sure they want you to show them that you know how to wade through the engineering gobbledegook instead.

However, an inspector/technician/craftsman would do it this way.

1) Use the known dimensions to determine the total volume of the shape. Since the problem gives only one outer diameter, it is safe to assume that the outer shape is a regular cylinder. Say the outer volume = OV.

2) "Using the above equation, I determined (Rt^2 - Rb^2) = 0.007340719 ft^2" So save this value. You will use it later.

3) Using the known weight and the known density of the brass, calculate the required dimensions of a hollow cylinder 15 inches long with an outer diameter of 8 inches.

d = 8

r = 4

a = pi x r x r = pi x 16

a = 50.2857 = approx. 50 sq. in.

OV = 50 x 15 = 750 cu.in. (or however many significant digits you need)

4) Now you are cooking with gas. You have a known volume of 239.62 cu.in. for the casting. Given the known OV, how thick will the walls need to be in order to reach the calculated volume of 239.62 cu.in.? Once you have that number, you can derive the AVERAGE inner diameter (AVID).

Simple enough. I know you know this already, but what the heck. I am bored. 239.62 / 15 = 15.97 sq.in. for the average area at the end of the inner cylinder.

15.97 / 3.14 = 5.09 diameter

5.09 /2 = 2.5 in. average inner radius.

5) Last, take your calculated difference between the top and bottom inner diameters. Add one half to the AVID to get the upper inner diameter, subtract one half from the AVID to derive the lower inner diameter. Or vice versa, whichever is larger. Works either way for this kind of problem.


This is the kind of ugly field shortcut that you will see surveyors, construction people, and inspectors using all the time. They tend to drive P.E.s crazy for some reason. But they work. Dunno if it will help you or not. But it's all I have to offer.

[CX]

There has to be a pre-written formula for calculating the circumference, diameter, slope, and weight /resiliency ratio for those breasts.

Yeah, well God definitely got all of that just perfect when he made her, 'cause dayumn....

Thanks, CX. You made me feel much less guilty about nixing my 16 year old daughter's plans to go to a live performance of Rocky Horror at midnight last night. She went to a party instead and was home by 11.

You've seen the movie, right? So, yeah, it's definitely not something I'd recommend for anyone under the age of 18. There were no male appendages, but there were some banana hammocks and some womens' underwear that didn't leave much to the imagination. Can't speak for the cast and the crowd where you are, naturally.

[CX]

"A vertical true centrifugal castig process is used to cast brass tubing that is 15 in. long and whose outside diameter = 8.0 in. If the speed of rotation during solidification is 1000 rpm, determine the inside diameters at the top and bottom of the tubing if the total weight of the final casting = 75.0 lb."

"The difference in inside radius between the top and bottom is related to the speed of rotation as follows: N = (30/Pi)[(2*g*L)/(Rt^2 - Rb^2)]^1/2
Where L = vertical length of the casting, Rt = inside radius at the top of the casting, Rb = inside radius at the bottom of the casting, g = the gravitational constant (32.2 ft/s^2), and N = the rotational speed in RPM."

Using the above equation, I determined (Rt^2 - Rb^2) = 0.007340719 ft^2

The density of brass = 0.313 lb/in^3, which I used to determine the volume of the cast part to be 239.62 in^3.

There is no illustration of the part, but none of the problems require them since the problems are all basically numbers games of manipulating the numbers you have to get the numbers you want.

[Caveat: The following all assumes that I am reading the question correctly, and that I have not gone senile. Two fairly large assumptions. ]


No doubt this is not the way they want you to derive it, but if you have the outside dimensions and the weight, plus the density, you have everything you need to just get the actual numbers. I am sure they want you to show them that you know how to wade through the engineering gobbledegook instead.

However, an inspector/technician/craftsman would do it this way.

1) Use the known dimensions to determine the total volume of the shape. Since the problem gives only one outer diameter, it is safe to assume that the outer shape is a regular cylinder. Say the outer volume = OV.

2) "Using the above equation, I determined (Rt^2 - Rb^2) = 0.007340719 ft^2" So save this value. You will use it later.

3) Using the known weight and the known density of the brass, calculate the required dimensions of a hollow cylinder 15 inches long with an outer diameter of 8 inches.

d = 8

r = 4

a = pi x r x r = pi x 16

a = 50.2857 = approx. 50 sq. in.

OV = 50 x 15 = 750 cu.in. (or however many significant digits you need)

4) Now you are cooking with gas. You have a known volume of 239.62 cu.in. for the casting. Given the known OV, how thick will the walls need to be in order to reach the calculated volume of 239.62 cu.in.? Once you have that number, you can derive the AVERAGE inner diameter (AVID).

5) Last, take your calculated difference between the top and bottom inner diameters. Add one half to the AVID to get the upper inner diameter, subtract one half from the AVID to derive the lower inner diameter. Or vice versa, whichever is larger. Works either way for this kind of problem.


This is the kind of ugly field shortcut that you will see surveyors, construction people, and inspectors using all the time. They tend to drive P.E.s crazy for some reason. But they work. Dunno if it will help you or not. But it's all I have to offer.

I ended up doing something very similar. Basically I calculated the volume of the cylinder using the weight and the density, then I figured out an average inner radius using the formula for the volume of a hollow cylinder. Technically in this type of molding operation, the cross-section of the inner surface is parabolic, but since this apparently isn't exact, I basically took that average radius and multiplied it by two (since the average = (Rt + Rb)/2), and since I already had (Rt^2 - Rb^2) = 0.007340719 ft^2, I basically solved both problems simultaneously by solving the second equation for Rt and substituting it into the first one. The answers I got looked valid given the other known dimensions, but I probably won't know for a while if I ended up getting it right or not.

[blacknblue]

Good luck then. Don't be surprised of your prof blows a gasket though.

[Distracted]

Thanks, CX. You made me feel much less guilty about nixing my 16 year old daughter's plans to go to a live performance of Rocky Horror at midnight last night. She went to a party instead and was home by 11.

You've seen the movie, right? So, yeah, it's definitely not something I'd recommend for anyone under the age of 18. There were no male appendages, but there were some banana hammocks and some womens' underwear that didn't leave much to the imagination. Can't speak for the cast and the crowd where you are, naturally.

I've never actually seen the movie...not really my thing. My husband has, though, and he was fully in agreement with keeping his 16 year old daughter out of that environment at midnight on Halloween. For some reason, she didn't understand our objections. Teenagers make me scratch my head. Why would she think that we'd allow her to go out until 2 am to see a play with nudity (and audience participation ) at 16? I never did stuff like that when I was that age.

Come to think of it... I've NEVER done stuff like that.

[leslina]

Thanks, CX. You made me feel much less guilty about nixing my 16 year old daughter's plans to go to a live performance of Rocky Horror at midnight last night. She went to a party instead and was home by 11.

You've seen the movie, right? So, yeah, it's definitely not something I'd recommend for anyone under the age of 18. There were no male appendages, but there were some banana hammocks and some womens' underwear that didn't leave much to the imagination. Can't speak for the cast and the crowd where you are, naturally.

I've never actually seen the movie...not really my thing. My husband has, though, and he was fully in agreement with keeping his 16 year old daughter out of that environment at midnight on Halloween. For some reason, she didn't understand our objections. Teenagers make me scratch my head. Why would she think that we'd allow her to go out until 2 am to see a play with nudity (and audience participation ) at 16? I never did stuff like that when I was that age.

Come to think of it... I've NEVER done stuff like that.

Because they're teenagers and they're entitled.

A typical weekday at work for me:

"Miss, do we gotta do work today?"
"Miss, why can't we just do nuthin' today?"
"Miss, why I gotta come to school? Why can't I just stay home?"

Needless to say, I fear old age if that's who's supposed to be taking care of me in the old folks home...

[JadziaKathryn]

Teenagers make me scratch my head. Why would she think that we'd allow her to go out until 2 am to see a play with nudity (and audience participation ) at 16? I never did stuff like that when I was that age.

Come to think of it... I've NEVER done stuff like that.

Ah, but that's never the point, is it? The point is that you are deliberately and cruelly trying to ruin her life for no discernable reason. *eyeroll* Hormones wreak havoc on common sense.

All things said and done, though, my parents would have been apoplectic if I so much as signalled an interest in such an event at 16. Actually, my mom would be apoplectic now if I so much as signalled interest in such an event, and were I interested I'm quite old enough to go.

[CX]

My parents were amused. Dad actually asked if I went home with someone else or if I came home by myself. My parents know me well, so he knew the answer to that one before he even asked.