UKH

Seems my brain is mush.

If I have a circle centre point C, radius R

and draw a tangent T out from point P (the bottom) a distance L to point A

and then a line of length T back to just touch the circle at point B

and <BCP is greater than 90o (i.e the contact point is above the midpoint of the circle).

What is the formula for the angle that line T makes with the tangent at point B?  (T is less than L but greater than L-R)

Assuming I've understood correctly:

The distance from the intersection of the tangents (call this point D) to the edge of the circle = r*tan(0.5 theta) Where theta is angle BCP.

BDA is a non-right-angled triangle. Side lengths T, rtan(0.5 theta) and L-rtan(0.5theta)

Using the Cosine Rule gives:

cos(ABD)=[T^2 + (rtan(0.5theta))^2 - (L-rtan(0.5theta))^2]/2*T^2*rtan(0.5theta)

This assumes you know angle BCP, the radius and lengths T and L.

Thanks... but there's only a single tangent, not two.

Imagine a stick leaning, in the plane of the wheel, against its rim so that it touches above axle height.

I can vary L (the distance the foot of the stick is from the base of the wheel along the ground); T is fixed (length of stick), as is R (radius of wheel), and I want to achieve a certain angle.

But I might have misunderstood your reply!

And I've just realised I've called two non identical lines T.  The first is L not T!

I don't understand what you mean by "the tangent at point B", the question implies you want the angle between this tangent and the stick. The other tangent is the ground. Is it the angle between the stick and the ground you're looking for?

No the angle between the stick and the wheel rim

Is it the angle you want, or the length L in order to achieve a certain angle?

In that case that's what I've given you I think. The angle between the tangent to the wheel rim at the contact point and the stick. The second tangent is the ground.

I think we need a diagram!

If you want the angle then:  theta = (arccos ((R^2-L^2)/(2*R*T)))  -  90

If you want the distance L then:  L^2 = R^2 - 2*R*T*Cos(90 + theta)

So you have 2 lines of equal length both starting from point A. The first reaches the wheel rim at a tangent. Surely therefore, because they're the same length, the second must do likewise?

I don't think they are equal length. Only L is a tangent. T is shorter than L if I've understood correctly.

Ah, easy mistake confusing the tangent T with the line of length T!

yeah, they're not identical length.

I'm pretty sure I have the right diagram in front of me but it's making my head hurt.

Correct. If it's too short, the stick falls on the ground; too long and it becomes a tangent

Oooh - That looks promising!  Thank you.

I will cut a stick and find a wheel and a protractor and report back

https://photos.app.goo.gl/fEmEL77M5L9pP8qV9

Is this the right diagram? Ignore the fact I've used different letters for some of the points. 

That's how I drew it

except you need to swap L and T

Yep, just realised that.  I have an answer but I don't like it.

https://photos.app.goo.gl/vCRQQbDuTVopEdCq7

Pretty sure my answer is correct

I'm now trying to persuade myself that your rather elegant answer is the same as mine...

Edit: bugger.  Your way was much better than mine.  That was a slightly wasted 20 minutes of my life.

Draw in the line OQ (on your diagram), then a bit of Pythagoras followed by Cosine rule should get you to the answer

Yep.  I am now annoyed at myself for going the long way around!  Possibly I shouldn't do maths after my second pint.

This simplifies to arcsin((L^2-R^2)/(2*R*T))

But it can't be right because to get theta=0 would require L=R when it clearly happens when L=T

I think the answer is arcsin((L^2-T^2)/(2*R*T))

Yes!

So now that I have stopped confusing the issue by duplicate naming one of the lines, has Robert cracked it?

And, if he has, could he rearrange as L=(function of theta) please?

Apologies, even though I have T^2 - L^2 written down in front of me several times, I somehow typed R^2 - L^2.  

As Robert says, arccos ((T^2-L^2)/(2*R*T))  -  90  is the same as arcsin ((R^2-L^2)/(2*R*T))

Thanks everyone - great job.

L^2= 2RTsin(theta) + T^2

I just used the cosine rule:

a^2 = b^2 + c^2 -2bcCos(A)

Where a^2 is Length C to A, and angle A is angle CBA.

this gives: L^2 + R^2 = c^2 (pythagoras)

and c^2 = R^2 +T^2 -2RT*Cos(x), where x is angle CBA

and x = 90-y say, where y is required angle

simplifying gives:

L^2 =T^2 -2RT*Cos(y-90)

therefore y =arccos ((T^2-L^2 )/2RT)+90.

now to look through everyone elses solutions!

Edit: just seen you can tidy this up by transforming cos(y-90) to sin(y).

So I get: y =arcsin ((T^2-L^2 )/2RT)

Is everyone else getting this too? OP, what’s the correct solution? My solution seems slightly different to everyone elses. 

Same as mine

Yours is L^2 - T^2, mine is the other way round. Can you see an error in my calcs or is it in yours? 

I think x=90+y, not 90-y

You’re absolutely right. Error in my calcs! Thanks for spotting.

so it’s: L^2 =T^2 -2RT*Cos(y+90)
and so it follows:

cos (y+90) =(T^2-L^2)/2RT

cos(y+90) =-Sin(y), therefore:

sin(y) = (L^2-T^2)/2RT, and henceforth:

y= arcsin((L^2-T^2)/2RT)

i get the same as you now. 👍