Where to plant the rose? The mathematics of gardening.

SydneyRoverP6B

SydneyRoverP6B

Well-Known Member
Staff member
You have just purchased a nice new rose and now you must decide where to plant it. The only available position is in a rectangular garden bed between a rock wall which is 5 metres high and your side fence which is 2 metres high. The straight line distance between the rock wall and the fence is 3 metres. How far from the rock wall do you plant the rose in order to ensure that the plant receives the most hours of sunlight on a day when the sun passes directly overhead?

Ron.
 
You need to be careful of the summer sun too, though. I planted one against a wal and the summer sun crisped it within weeks. :cry:
Best bet would be to put it in a pot and shift it around, alternatively stick it in a box and send it to Cape Town, I will look after it, I promise!! :LOL:
 
I have to agree

Where's the fun in trying to work out where to plant it. It'll only defy logic and do its own thing anyway :roll:

Just stick it in the ground and be done with it.

Dave
 
It is not a serious gardening question, rather I just dressed up some exercise for the brain with a rose... :D

Good answers nonetheless,...albeit not the right ones... :D

Ron.
 
So what thoughts do you have as to how the problem might be tackled? No matter how complicated a problem might appear on the surface, all can be broken down into a series of simple steps so as to form a solution.

Ron.
 
Havent had a chance to take a proper look at this one next.

I'm guessing the ideal point would be where the intersection is between the top of the left side and the bottom right and the top of the right side on the bottom left. the red line in the attached jpeg is what we're looking for?

Rich.
 

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Hello Rich,

If you have made a scale drawing, then you have assumed that the plant is a certain height in order for it to be a point of intersection with the lines that you have shown. As no height was given, you can't assume that. Imagine that the plant has zero height to start and work from there.

Ron.
 
Hi - thanks for the tip :O) Saves me going off on another 'tangent' pardon the pun ;o)
 
Hello Rich,

I'll give you a further pointer. Look carefully at what you wrote in your last post and think about an angle, but the reciprocal.....will lead to the way.

Ron.
 
so given what you've said i think we're looking at a diagram more like this?
 

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Hello Rich,

Your second diagram is better but,....ok...now connect the top of the 5m line to the X you have on the 3 metre line. Connect the top of the 2 metre line also to the X. Now right an expression for the angle subtended by those two lines that I just asked you to draw in. Better still,...I'll give you the expression and you show me how it is derived.

Angle theta = PI - arccot(p/5) - arccot((3 - p)/2),...where theta is the angle between the two lines I just asked you to add, while 'p' is the location of the X. This is just the first part, once you have derived that expression we'll go from there.

Ron.
 
Hello Rich,

How are you travelling so far?

Richard,..Darth,..?

Who else has some thoughts that they might like to share?

Ron.
 
Sorry - i've been a bit tied up changing the engine over the weekend - I've not forgotten - when i've got a spare half hour i'll be on it again ;o)

rich.
 
SydneyRoverP6B said:
Hello Rich,

How are you travelling so far?

Richard,..Darth,..?

Who else has some thoughts that they might like to share?

Ron.
Click to expand...

Been a bit busy Ron - with this
NSRb.jpg


I can only assume that the best place to plant it would be 5 parts from the 5M wall, and 2 parts from the 2M wall, i.e.

(5/7) * 3 = 2.14M from the 5M wall

Richard
 
Hello Rich and Richard,

No worries gents,..I was just wondering how you were both going.

Gee Richard,... :shock: That looks pretty grim,..lucky you're a handy fellow and you have the abilities to put that right.

quattro wrote,..
I can only assume that the best place to plant it would be 5 parts from the 5M wall, and 2 parts from the 2M wall, i.e.

(5/7) * 3 = 2.14M from the 5M wall
Click to expand...

Sorry Richard,..if only it was that straight forward.

Ron.
 
i'm pretty convinced that the secret to this is to work out the length ofthe line that goes from the bottom of the tall wall scrapes the 2m wall then to the point where it meets the circle which the sun travels accross. Will have a proper look in my lunch hour:)
 
Well the equation you've provided comes from tan angle1 = 5/p and tan angle2 = (2/3-p) and inverted and minused from PI to give the remaining angle between the 2 lines eminating from X...
 
from here you could obviously graph it to find the maximum value of theta given all values of P - seems like you're likely to have a nicer solution in mind though!

That gives theta as about 0.52 rads then we can solev for P. Would be interested in knowing your better method for getting a max value of theta...

Rich
 
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