# Boat crossing/Glide Ratio/Hydraulic design

1. Jan 19, 2013

### zeshkani1985

Ok I had these 3 questions: I believe I have ansered the first 2 but I' am just stuck on the third.
1. an aircraft has a glide ratio of 12 to1 m, a building that is 45m high lies directly in the glide path, the plane clears the building by 12m, how far is the run way from the building?
ANS: I just took the 12mx45m= to give me 540m i' am pretty sure this is right?

2.to cross a river that is 1km wide, with a current of 6km/hr, a boat that is perpendicular to the far bank, the boat speed is 10km/hr. at what position will the boat actually contact the far bank? what direction should the boat have to cross directly across?
ANS: here i made a right triangle and just used the 6^2 +10^2 = x^2, so he will actually contact the far bank at 11.6km away. 2part: since having figured out his actual contact point i figured the angle and by using the law of sines that it was 31.3 degrees
I think the first part is right, i' am not sure about the 2nd

3. design a hydraulic holding system for a hay-baling system, the system has four cylinders with a 120mm diameter pistons with a stroke of 0.320m, the lines connecting the system are 1cm id(inside diameter). there are 15.5m of lines in the system. for proper design, the reserve tank should hold a minimum of 50% more then the amount of hydraulic fluid in the system, if the diameter of the reserve tank is 30.48cm what is the shortest height it should be?

This question I' am completely stuck on, I have no idea were to start, any pointers would be great

2. Jan 19, 2013

### rude man

Not quite. You missed the bit about the plane clearing the building by 12 ft.

3. Jan 19, 2013

### zeshkani1985

so it would be 540m-12m = 528m ?

4. Jan 19, 2013

### rude man

You mean, the faster he goes towards the other bank, the longer it would take him? Like, if his speed is 100 km/hr he would land 100^2 + 6^2 = x^2 below where he started? You know that can't be right.

Hint: he takes the same amount of time to reach the other shore no matter how fast the river is moving ... so how much time is that, and then how far below his starting point does the flowing river take him?

5. Jan 19, 2013

### rude man

No. Doesn't it make sense that the higher he clears the building, the farther he has to fly before hitting the runway? Draw a picture ...

6. Jan 19, 2013

### zeshkani1985

okay i think i got it, so the plane is acutally 12m above the building, if this is the case, then there is an extra 12m

7. Jan 19, 2013

### zeshkani1985

okay since the boat velocity is at 10km/h, and the river at 6km/h
6km/h __
| /
10km/h|/ since there are 2 velocities, 102+62=11.6km/h thats the diagonal speed
so with this i can just divide the distance of 1km/10km/h =t to get time and multiply velocity by time to get the distance 11.6km/h x 0.1h= 1.16km is this right ?? for the first part

8. Jan 19, 2013

### zeshkani1985

for the second part in question 2

can i just take the inverse of cos of the distance of the river/by the boat speed

cos=1/10=85 degrees

9. Jan 19, 2013

### rude man

You have the total distance traveled right but the question really asks how far below the launching point does he arrive on the other side.

10. Jan 19, 2013

### zeshkani1985

11. Jan 19, 2013

### rude man

No.

He'll point his boat at an angle θ from the straight-across direction. So write an equation giving the along-the-bank distance he travels in time T. You can set up a coordinate system fixed to the ground, with the river flowing in the -y direction and the distance along the river bottom = x. x = 1km and y = 0 would be the point directly across from his launch point, which would be x = y = 0.

Last edited: Jan 19, 2013
12. Jan 19, 2013

### rude man

13. Jan 19, 2013

### zeshkani1985

i just looked at the problem again and i set it up differently and i took the inverse tan
and I got a 5.7 degrees, but since this is just the angle, would you add the 180+5.7 to get 185.7 ?

14. Jan 20, 2013

### rude man

I don't know where you got the inverse tangent.

Set up as I suggested and write the distance equation for the y coordinate. You don't know T, the time of crossing, but you don't need to to answer part 2.

(The angle is between 0 and 90 degrees, that should be obvious. And it's bigger than 5.7 deg.)

15. Jan 20, 2013

### zeshkani1985

okay i did set it up how you said so here is my question, i' am just kind a really confused over it

i will be using sin, since i know the opposite and hypotenuse since the opposite it 6km/h for the hypotenuse do i use the 10km/h or the 11.6km/hr from the previous answer if i just use the 6/10 i get 36 degrees if i use the 6/11.6 i get 31 degrees

16. Jan 20, 2013

### zeshkani1985

i think i got it the angle should be 31 degrees i used this and back substituted it and found out that is had to b e 31 degrees.

17. Jan 20, 2013

### rude man

What equation did you use to arrive at 31 deg? If you did as I suggested you first got an equation for the total y component of velocity of the boat wrt the ground (riverbottom). This equation includes the y component of the boat's intrinsic speed (speed on a quiet surface) plus the y speed of the river. You then multiplied this resultant y speed by time T to cross the river to get the final y position. You then set that final position to y = 0.

The y component of intrinsic boat speed is going to include the angle he points the boat to.

18. Jan 20, 2013

### zeshkani1985

okay here i did it again

since i know that his displacement was .60km due south and the river is 1km wide

so to balance it out he had to go .60km due north to cross directly and i took the inverse tan of .60km/1km to get 30.96 which is about 31 degrees

i think i might just give up