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scr1
3rd Jan 2015, 13:59
Help

Son has come to me for help with his trigonometry homework. I have helped him through most of it but I am stuck on one question. ( it is over 25 years since i did this). Can any one please help on how to do this.

the question is

A vehicle has a wheelbase of 3.0m and a track of 1.8m. When the vechicle is turning through a certain radius, the steering angle of the inner wheel is 25 degrees.

Calculate using trigonometry

The length of radius of turn (measured to the center of rear axle)

The angle of the outer wheel

Thank You

Bushfiva
3rd Jan 2015, 14:04
Draw lines perp to the wheels to the center of the circle. You know the angle between the lines at the center. You know they're the same length. You know the wheelbase. Everything else follows quite simply.

fitliker
3rd Jan 2015, 14:04
Radial or cross ply tires ?

joy ride
3rd Jan 2015, 14:18
The trigonometry can provide one answer, but full details of the vehicle in question might make it wrong! If it is a 3 wheeler it may turn in a tighter radius, if it has 4 or more back wheels, or two or more back axles then the radius would be bigger. If the vehicle has a solid pivoting front axle or other non-Ackerman steering geometry it would also vary.

Mr P. E. Dant

UniFoxOs
3rd Jan 2015, 14:23
Fitliker, I suspect as this is a trig question the slip angle is not going to be taken into account. Therefore the radius of the circle at the front wheels is going to be easily calculated (drawing suitable triangles to sort out what formula to use), and the position of the centre of the circle located relative to the vehicle.

The rear wheels get a bit more difficult but it must be turning around the same centre (otherwise it would be over- or under-steering), it will just be describing a slightly larger radius. Therefore a bit more drawing of triangles to work out which formula to use will be needed.

Checkboard
3rd Jan 2015, 14:28
Radius of turn 7.33 meters?

The angle of the outer wheel is 20.1 ?


This is ignoring slip angle (the difference between the angle the tyre is pointing and the direction of travel - a tyre is always skidding in a turn).

RJM
3rd Jan 2015, 14:30
I take it you're assuming the earth is flat? Spherical trigonometry isn't that hard...

Fox3WheresMyBanana
3rd Jan 2015, 15:07
Checkboard is correct. Working as follows:

http://www.atu587.com/sites/default/files/Cornering%20Forces%20and%20Geometry.pdf

See slide 2 for the diagram

Let inner front wheel be A, inner rear wheel be B and centre of circle be C
angle ACB = 25 degrees also.
Opposite side (AB) = wheelbase = 3.0m
Therefore adjacent side = distance from inner rear wheel to centre (BC) = 3.0/tan25 = 6.4335m

Length of radius of turn = BC+half track = 6.4335+0.9 =7.3335m

Let outer front wheel be D, outer rear wheel be E and centre of circle be C

Opposite side = wheelbase = 3.0m
Adjacent side = BC+track = 8.2335m
So angle of outer wheel = tan^-1(3/8.2335) = 20.02 degrees

boguing
3rd Jan 2015, 15:19
Checkboard.

Agree on radius, but I get 20.02 degrees?

boguing
3rd Jan 2015, 15:20
Beaten by F3WMB.

Fox3WheresMyBanana
3rd Jan 2015, 15:24
Checkboard gets 20.1 by rounding to 8.2 before doing the inverse tan.
It is advisable to keep one extra significant figure than required in the answer for intermediate stages of calculation, and this is what the teacher will expect.
Mostly, 3 sig.figs is required in a final answer, so 4 sig figs should be used for the inverse tan. Because the final figure was a 5, I used 5 sig figs to be sure.

Answers should be given as 7.33m and 20.0 degrees.

scr1
3rd Jan 2015, 15:37
Thank you for all your help, think i understand it now

wings folded
3rd Jan 2015, 16:16
Would it now be safe to be a passenger in your car when you turn a corner?

Checkboard
3rd Jan 2015, 16:36
I was solving it in my head ;)

The calculations were done on an online calculator without a memory, so I was just memorising intermediate numbers. Sorry for the error. :)

funfly
3rd Jan 2015, 16:57
Angle of the outer wheel will be the same as the inner wheel.

They will not 'follow the curve' as both are turned by the steering mechanism so one will slip a bit as the turn progresses.

Checkboard
3rd Jan 2015, 18:46
funfly: https://en.wikipedia.org/wiki/Steering


https://upload.wikimedia.org/wikipedia/commons/a/a0/Ackermann_turning.svg

Fox3WheresMyBanana
3rd Jan 2015, 19:06
The length of radius of turn (measured to the center of rear axle)

from the OP

rh200
3rd Jan 2015, 20:27
Funfly, google "toe out on turns".

What will the angles be if its a twin steer?:p

G-CPTN
3rd Jan 2015, 20:34
The length of radius of turn (measured to the center of rear axle)
A case of 'read the question carefully'- otherwise the answer could be wrong for the sake of simple arithmetic.

ExSp33db1rd
3rd Jan 2015, 23:29
I trust you used the correct turn signal before deviating from the straight line ?

parabellum
3rd Jan 2015, 23:43
Reminds me of the trig we did at school, "A man places a twenty foot ladder against a window that is twelve feet from the ground, how far is the base of the ladder from the wall?" My answer of, "Who gives toss? but DO make sure that either the top of the ladder is securely fastened to the wall or that you have a person holding the bottom of the ladder" got me a short spell standing outside the headmaster's office followed by three of the best and a sore bum! :)

parabellum
5th Jan 2015, 03:40
Standing on bottom rung rather than just holding - OK, give you that one Henry but definitely a big no to just securing the bottom, insufficient, the top can still slide left or right.

probes
5th Jan 2015, 06:42
Awesome! (was this the most irritating word the 'murricans use? :))

Now - show the thread to the teacher so that he/she'd know what the kids were supposed to do, with real life entering trigonometry :cool:.

MagnusP
5th Jan 2015, 07:30
And why would you rest a ladder on the window anyway? Surely it should be on the wall adjacent to the window or the cill below. Seems a damned dodgy way to run a windowcleaning business.