Recent content by trumpsnuffler

sorry, still getting to grips with using this forum. t1 and t2 were meant to be written in subscript..

oh wow. no idea why i didn't think of that.. :blushing: okay. so cos² = 1 - sin² cos² = 1 - sin² = 1 - [(n/v)sin(φ)]² = 1 - (n²/v²)sin²(φ) therefore cosθ = √[1 - (n²/v²)sin²(φ)] is the above expression the value of [t][/1] then? and [t][/2] = the same, but a positive...

okay. as (n/v)sin(φ) = sin(θ) (n/v)cos(φ) = cos(θ) wasn't too sure if i could make this jump, but bearing in mind that the graphs of sin and cos are so similar, surely the above expression for cos(θ) holds to be true?

and with regards to the second part of the question 'find the maximum range and for which direction', would i have to go through the normal means of finding a maximum (i.e differentiating), and if so, differentiating with regards to what? or do i just have to make n²sin²φ = 0, which would give...

morning all, opted to get an early night in favour of an early rise! okay, the above vector diagram's a bit misleading. on the return journey, the plane doesn't necessarily come fly off westwards to form a right angled triangle with Q v and n. right the x and y components for the...

hmmm, okay, would the time be calculated via pythagoras's theorem for the individual x and y components of v and n, for each journey repectively? (n/v)sin(φ)..? i'm a bit lost, i can see why the numerator is R(v+n)(v-n) and i can see that the v on the denominator comes from R/v, but i...

hmm. what is this new variable S...? (sorry to be so slow, please be patient with me!)

i think at the rate this question is taking me, you might break 15,000 posts by the end tim...

so would this be the (v + n) from the journey there, and then (v - n) for the journey back? or (v - n) for the way there, and (v + n) for the way back (now i know that a north wind blows north and not south...)

the vectors leading to Q? i.e. v and n?

hmmm. latex doesn't seem to work for me..

isn't it just the same? the angle made from the north axis to the resultant vector Q (in the return journey) must be φ in order for the plane to return to its original place. so, from this, and the 'z rule' of angles, the top angle, between n and Q must therefore be equal to φ. so...

sorry for the delay in a reply, had to drop my housemate off to pick up his car. okay, so from the sine rule, i get: sin (180-φ)/v = sin(C)/n and from this, so C = arcsin(nsin(180-φ)) <-------- can this be canceled to give n(180-φ)? and then the final angle is just 180 - (180 - φ) - the...

hmmm, I'm not too sure I'm either a) working out the angles correctly, or b) working out the correct angles.. but okay. the angles i got for the internal angles of triangle vQn are: the angle between n and Q is (180 - φ) then the angle at the top, called, say, A, between v and n...

hmmm, I'm not too sure I'm either a) working out the angles correctly, or b) working out the correct angles.. but okay. the angles i got for the internal angles of triangle vQn are: the angle between n and Q is (180 - φ) then the angle at the top, called, say, A, between v and n, through use...