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View Full Version : ok, see what y'all think about this.


Sei'taer
07-31-2008, 08:38 AM
We recently changed our standards on sidewalk to match the city of Memphis. Now, maybe I'm just terrible at explaining this kind of thing...I'm really not sure. I was going to explain it on here and see what y'all thought about it. I'll just have to give you a little bit of an idea what I am talking about because most of you have probably never noticed this or even know what it is. So, here we go:

All sidewalks have slope. Normally they slope to the street. This way, the water runs downhill and goes into the street to drain away instead of pooling up on the sidewalk. If there is grass between the sidewalk and the street, then the water drains into the grass and some seeps in and some goes on to the street and drains away. We recently changed our standard to make the sidewalk slope 1/4" per foot. Our sidewalks are 5' wide with a 4' 6" grass strip. The slope across the grass is 1/2" per foot. So all you do is take 1/4 * 5'and then take 1/2" * 4'6" and it gives you the total slope from the back of the curb to the back of the sidewalk...then all you have to do is set everything up to match that slope. I think this is pretty easy, because in measurements like this I can relate it to money and do it in my head (funny how that works).

So the question is, what is the correct total slope for the 9'6" and why is it so hard for people who actually do this kind of work to understand? I don't get it...but maybe I'm making more out of it then I should. BTW, here is the standard detail for a sidewalk/driveway apron (http://www.memphistn.gov/engineeringforms/38_ResidentialDriveway.pdf)...it gives you all the same measurements I did so if it's helpful to look at it feel free. It's a pdf just so you know.

Brita
07-31-2008, 08:54 AM
So....the slope for the sidewalk would be 1 1/4" and the slope for the grass apron would be 2 1/3" (this one I'm not sure on, metric is so much easier)?

Crispin's Crispian
07-31-2008, 10:24 AM
So....the slope for the sidewalk would be 1 1/4" and the slope for the grass apron would be 2 1/3" (this one I'm not sure on, metric is so much easier)?

I get 3 1/2" total from the back of the sidewalk to the curb.

5' * 1/4" = 1.25" or 1 1/4"
4'6" = 4.5'
4.5' * 1/2" = 2.25" = 2 1/4"
Total = 3 1/2" from the property side to the street side.

So if you have to have a 6" curb,, the high point of your sidewalk has to be 9 1/2" above height of the street.

1/2" per foot seems pretty steep. But them I'm terrible at imagining what that would feel like.

Brita
07-31-2008, 10:28 AM
Do you combine the slopes or grade each section seperately?

Ivhon
07-31-2008, 11:11 AM
Is this gonna be on the GRE?

GonzoTheGreat
07-31-2008, 11:42 AM
Why aren't there different units for pavements and grass? For that matter: why use the same units for horizontal and vertical directions? It seems to me that with just a tiny little bit of effort, the system could be a whole lot less rational.

Davian93
07-31-2008, 11:52 AM
In Soviet Russia...Car drive you.

Sei'taer
07-31-2008, 12:46 PM
Do you combine the slopes or grade each section seperately?

On the sidewalk, you have 2 seperate grades. On the driveway apron, you have one straight grade throughout...from front to back.


Sdog had it right, but it's much easier to check if you say 3 1/2" above the top of the curb instead of 9 1/2" above the street...normally the 6" curb is a given, unless you have something strange like a rolled back or mountable curb (this is sloped so that it can be driven over...instead of being straight up and down like normal curb...although normal curb isn't straight up and down either, buts thats for a different time).

1/2" per foot seems pretty steep. But them I'm terrible at imagining what that would feel like.

Also SDog, most places set the driveway apron grade to right around a 1/2" per foot. Ours is at about 3/8" per foot (give or take a little). SOme of the interstates are also graded at a 1/2" per foot so that the water drains off faster and because of the wider roads/higher speeds it's really not noticeable until you get caught in a traffic jam...and even then you have to know what you're looking at.

Why aren't there different units for pavements and grass? For that matter: why use the same units for horizontal and vertical directions? It seems to me that with just a tiny little bit of effort, the system could be a whole lot less rational..

Watch out, you're getting close to survey talk.

Brita
07-31-2008, 12:57 PM
Sdog had it right,

~~sigh~~ well, at least my math skills are decent enough to figure out BSA, chemo doses and dose reductions. Otherwise I'd really be in trouble :o

Davian93
07-31-2008, 12:59 PM
~~sigh~~ well, at least my math skills are decent enough to figure out BSA, chemo doses and dose reductions. Otherwise I'd really be in trouble :o

Yeah...that could get dangerous fast...;)

pops taer
07-31-2008, 01:26 PM
An 80 lb dog at 2.2 lbs per kilo weighs in at, round off for ease, approximately 40 kilos. If the dog is more than 10 percent dehydrated...the initial assumption is to begin fluid therapy anticipating a replacement of approximately 4 liters...Is that right Brita????

Sarevok
07-31-2008, 01:54 PM
uhm... i doubt that, since it would mean the dog is 100% water....

JSUCamel
07-31-2008, 02:35 PM
lmao sare

Brita
07-31-2008, 02:42 PM
LOL pops! 4 litres would be quite the bolus :)

I do simple calculations like:

Height 155 cm, weight 65 kg.
square root [(height x weight)/3600]= BSA, so BSA is 1.67m2 (we have online calculators that do this for us ;) )

Patient ordered Paclitaxel 200 mg/m2. So patient receives a dose of 334 mg of Paclitaxel.

Patient develops severe neuropathy, protocol calls for 25% dose reduction, so the next cycle (if no major weight change) the patient will receive 251 mg of Paclitaxel.

And I have no idea if the same principles of chemotherapy delivery apply to dogs :p

GonzoTheGreat
07-31-2008, 03:54 PM
pops, your estimate of 40 kilogram for 80 lb is about 10 percent too high (as should be obvious from the 2.2 lb per kg, instead of 2 lb). Furthermore, it is more reasonable to assume that a dog (or another animal) is something like 80 or 90 percent water, instead of 100% (as Sare points out). This means that the ordinary water amount in an eighty pound dog would be something in the range of 28 to 32 kilogram, and one tenth of that is about 3 kg. Since, as you correctly remembered, a liter of water weighs 1 kg, this means that you would need to insert 3 liter into the dog.

Alternatively, you can avoid getting the dog that far dehydrated. I think that would be the better option, but I'm not a veterinarian.