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Hi triical i need help again on angles
a straight line is an angle of 180 degrees
is there a way to know how wide a angle is?
If the box were 5 feet in length would i be able to tell how wide a angle would be directly in the middle?
not sure if you are combining 2 examples or is the wall 5 ft from the weapon?
No the weapon is on the wall which is 5ft wide and only sprays 40 degree angle , and yes i am combining two examples
hmm no its more like i will draw your same example
is it a top down, or birds eye view? |dw:1433116747233:dw| the circle is the top of the person's head
Birds eye i would say , its really a sprinkler i just want to check if i put this on a 40 degree angle to rotate would it hit the walls on each side or miss them?
I think it depends on how far the water can shoot out
Thanks for your help, i have many questions . Uhm lets just say its 35 feet long walls , would i be able to calculate that?
do you know how far the water can shoot out? the max distance possible
my bad is shoots up to 38 feet but i can adjust it to 35 and the walls are 35 feet long
sounds like you are going to consider the water pressure next
you don't really need to use trig or anything like that you just need to make sure that the distance (shown below) is more than 38 feet to keep the wall dry |dw:1433117166966:dw|
any other segment will be longer than 38 feet (assuming the black segment marked above is 38 feet) so if the black segment is say 40 feet, then the other segments are longer than 40 feet
|dw:1433117235149:dw| This is what it looks like
The water sprays from the point
I see now
where is that sprinkler point placed in relation to the 5 ft wall? At the halfway point?
yes, which makes me think it wont hit the walls but i have to make sure
directly next to the wall maybe 2 inches away
ok against the wall, but where along the wall?
directly at the 2.5 length
oh ok, thx
here i will draw what i have so far
Thanks for the help btw
anyways, since the sprinkler is at the halfway point, we know this |dw:1433117543177:dw|
I got the 34.833 from the fact that (35 ft ) - (2 inches) = (35 ft) - (0.167 ft) = 34.833 ft
pull out this triangle |dw:1433117745200:dw|
use the pythagorean theorem to solve for x
oh wait I'm not thinking one sec
yes cus 40 degrees
I'm going to assume that the sprinkler is aimed in a way such that these angles (marked below) are congruent. Put another way, the sprinkler is watering each half equally |dw:1433118000730:dw|
if that is the case, then each of those angles are 70 degrees since 70+70+40 = 180
i just noticed theres no sprinklers that start in the middle and go left and right grr
the water sprays at most 38 feet so the longest this segment (marked below) can get is 38 ft |dw:1433118100606:dw|
The question is: is the marked segment below 2.5 feet? |dw:1433118160441:dw|
IF the last segment marked was 2.5 ft, then a triangle would form |dw:1433118194993:dw| and this triangle would have its segment lying right against the wall
No its just if the water will ever hit the walls is my question
I know. What I'm trying to say is that IF this horizontal segment of the triangle is 2.5 ft, then the triangle will lie along the right wall. Which means the water will hit the wall IF the horizontal segment is longer than 2.5 ft, then the water will hit the wall but will come much closer to the sprinkler
I am still confused, how would you be able to check, i know if bullets went straight forever theyd eventually hit a wall but the water only goes 38 feet. and dies out afterwards that length
One moment. I'll try to paint a better picture
Sorry im not very comprehensive lols
and if i spelled comprehension in the wrong way i meant my comprehensions skills are below 0
I'm making a pic at the moment. one sec
I'm making a geogebra applet so you can see an accurate diagram
Thanks so much! What is an applet?
well ideally there is some level of interactivity, but I don't think that's needed here. Anyways, that would take longer than normal and I'm taking a long time as it is. So I'll just post a pic of what I have so far (see attached)
holy crap so it hits the wall nearly automatically
anyways, in that diagram, the distance from S to P is 2.5 ft using a bit of trig, we can figure out that the distance from S to K is roughly 7.31 ft but the water can shoot as far as 38 ft, so the water will definitely hit the left & right walls
yes it might be better to have the sprinkler set up halfway along the 35 ft wall then again, nvm it will still hit a wall since that 5 ft gap isn't that big
lols yeah ! Well i am very grateful for your help! Thank you! And Thanks again!
are you able to check at what degrees it doesnt hit the walls?
you're welcome, I suggest you check out geogebra. It's not as great as something like autocad, but it allows for (somewhat) quick scale drawings to figure out the distances/lengths easily. And the program is free, so another bonus.
Ok i sure will, thatll be helpful for this project, autocad is somewhat like autodesk inventor?
I've never heard of that other program, but perhaps? Autocad is mainly for use if you want to build blueprints or schematic drawings of objects (often in 3D).
Oh ok, i saw auto cad in my engineering class just never used it. Well i appreaciate your help, i have many other questions but i think i should close this one