If I’m driving a car at 60 km/hr and suddenly make a left 90 degree turn on hard lock, turning circle 11 metres.

How fast am I going at the end of the turn?

A) 60 km/hr (from conservation of energy)
B) 0 km/hr (from conservation of momentum)
C) Other speed
D) Not enough information
E) Can’t be done, the wheels will skid.

mollwollfumble said:

If I’m driving a car at 60 km/hr and suddenly make a left 90 degree turn on hard lock, turning circle 11 metres.

How fast am I going at the end of the turn?

A) 60 km/hr (from conservation of energy)
B) 0 km/hr (from conservation of momentum)
C) Other speed
D) Not enough information
E) Can’t be done, the wheels will skid.

Are we assuming a typical spherical blackbody car in a vacuum? I’m not sure it is physics otherwise.

mollwollfumble said:

If I’m driving a car at 60 km/hr and suddenly make a left 90 degree turn on hard lock, turning circle 11 metres.

How fast am I going at the end of the turn?

A) 60 km/hr (from conservation of energy)
B) 0 km/hr (from conservation of momentum)
C) Other speed
D) Not enough information
E) Can’t be done, the wheels will skid.

If there is zero friction, conservation of momentum requires that the car will continue at 60 km/h in the original direction, although I don’t know how the car would have got to 60 km/h in the first place.

If there is sufficient friction to allow the car to turn, then momentum will be transferred to the Earth, so that doesn’t help us.

If the engine is supplying sufficient energy to overcome energy losses, the speed will be 60 km/h.

If not, it could be anything.

Not enough information.

D

What type of tyres and is traction control off or on?

dv said:

Not enough information.

Yeah this.

The Rev Dodgson said:

mollwollfumble said:

If I’m driving a car at 60 km/hr and suddenly make a left 90 degree turn on hard lock, turning circle 11 metres.

How fast am I going at the end of the turn?

A) 60 km/hr (from conservation of energy)
B) 0 km/hr (from conservation of momentum)
C) Other speed
D) Not enough information
E) Can’t be done, the wheels will skid.

If there is zero friction, conservation of momentum requires that the car will continue at 60 km/h in the original direction, although I don’t know how the car would have got to 60 km/h in the first place.

If there is sufficient friction to allow the car to turn, then momentum will be transferred to the Earth, so that doesn’t help us.

If the engine is supplying sufficient energy to overcome energy losses, the speed will be 60 km/h.

If not, it could be anything.

Accelerator off. Traction control, either on or off.
Is there a big difference in final speed between turning while in gear and turning while in neutral?

My reason for asking is that conservation of energy suggests 60 km/hr but I’ve noticed while driving that it’s much closer to 0 km/hr. So steering can cause a massive energy loss, almost as much as braking. But why? I can’t seem to be able to draw force diagrams that make sense.

One thing that is clear is that because both front wheels turn the same angle, the perpendicular to both meets at infinity, not at the centre of the turning circle. This forces the front tyres to slide sideways, causing an energy loss.

Is this the dominant energy loss? Or is it the sliding sideways of both front and rear tyres under centrifugal action? How to calculate friction of the tyres on the road?

Physical testing in a real car?

I think you can see the turning losses like braking.

think back to when you were a kid, various bikes or whatever, that didn’t have brakes, you’d use turning for braking, in fact as emergency braking it features in your mind as a contingency should your car brakes fail.

mollwollfumble said:

The Rev Dodgson said:

mollwollfumble said:

If I’m driving a car at 60 km/hr and suddenly make a left 90 degree turn on hard lock, turning circle 11 metres.

How fast am I going at the end of the turn?

A) 60 km/hr (from conservation of energy)
B) 0 km/hr (from conservation of momentum)
C) Other speed
D) Not enough information
E) Can’t be done, the wheels will skid.

If there is zero friction, conservation of momentum requires that the car will continue at 60 km/h in the original direction, although I don’t know how the car would have got to 60 km/h in the first place.

If there is sufficient friction to allow the car to turn, then momentum will be transferred to the Earth, so that doesn’t help us.

If the engine is supplying sufficient energy to overcome energy losses, the speed will be 60 km/h.

If not, it could be anything.

Accelerator off. Traction control, either on or off.
Is there a big difference in final speed between turning while in gear and turning while in neutral?

My reason for asking is that conservation of energy suggests 60 km/hr but I’ve noticed while driving that it’s much closer to 0 km/hr. So steering can cause a massive energy loss, almost as much as braking. But why? I can’t seem to be able to draw force diagrams that make sense.

One thing that is clear is that because both front wheels turn the same angle, the perpendicular to both meets at infinity, not at the centre of the turning circle. This forces the front tyres to slide sideways, causing an energy loss.

Is this the dominant energy loss? Or is it the sliding sideways of both front and rear tyres under centrifugal action? How to calculate friction of the tyres on the road?

Physical testing in a real car?

One thing that is clear is that because both front wheels turn the same angle, the perpendicular to both meets at infinity, not at the centre of the turning circle. This forces the front tyres to slide sideways, causing an energy loss.

Steering geometry set-ups ensure that front wheels don’t turn the same angle.

Michael V said:

mollwollfumble said:

The Rev Dodgson said:

If there is zero friction, conservation of momentum requires that the car will continue at 60 km/h in the original direction, although I don’t know how the car would have got to 60 km/h in the first place.

If there is sufficient friction to allow the car to turn, then momentum will be transferred to the Earth, so that doesn’t help us.

If the engine is supplying sufficient energy to overcome energy losses, the speed will be 60 km/h.

If not, it could be anything.

Accelerator off. Traction control, either on or off.
Is there a big difference in final speed between turning while in gear and turning while in neutral?

My reason for asking is that conservation of energy suggests 60 km/hr but I’ve noticed while driving that it’s much closer to 0 km/hr. So steering can cause a massive energy loss, almost as much as braking. But why? I can’t seem to be able to draw force diagrams that make sense.

One thing that is clear is that because both front wheels turn the same angle, the perpendicular to both meets at infinity, not at the centre of the turning circle. This forces the front tyres to slide sideways, causing an energy loss.

Is this the dominant energy loss? Or is it the sliding sideways of both front and rear tyres under centrifugal action? How to calculate friction of the tyres on the road?

Physical testing in a real car?

One thing that is clear is that because both front wheels turn the same angle, the perpendicular to both meets at infinity, not at the centre of the turning circle. This forces the front tyres to slide sideways, causing an energy loss.

Steering geometry set-ups ensure that front wheels don’t turn the same angle.

tyre companies would love it if they did turn at the same angle.

transition said:

I think you can see the turning losses like braking.

think back to when you were a kid, various bikes or whatever, that didn’t have brakes, you’d use turning for braking, in fact as emergency braking it features in your mind as a contingency should your car brakes fail.

Speedway motorcycles slow for corners by using full throttle, laying the bike down and sliding the rear wheel. (Turn the brain off and throttle on is the uncomfortable mantra.) But this is not a significant component of molly is asking about. It could be a minor component.

Interestingly, modern motorcycle road and racing tyres allow lean angles well beyond 45° from vertical. My road-bike tyres have published lean angles of 58° in the dry and 53° in the wet! MotoGP riders often have lean angles beyond 63°.

Bogsnorkler said:

Michael V said:

mollwollfumble said:

Accelerator off. Traction control, either on or off.
Is there a big difference in final speed between turning while in gear and turning while in neutral?

My reason for asking is that conservation of energy suggests 60 km/hr but I’ve noticed while driving that it’s much closer to 0 km/hr. So steering can cause a massive energy loss, almost as much as braking. But why? I can’t seem to be able to draw force diagrams that make sense.

One thing that is clear is that because both front wheels turn the same angle, the perpendicular to both meets at infinity, not at the centre of the turning circle. This forces the front tyres to slide sideways, causing an energy loss.

Is this the dominant energy loss? Or is it the sliding sideways of both front and rear tyres under centrifugal action? How to calculate friction of the tyres on the road?

Physical testing in a real car?

One thing that is clear is that because both front wheels turn the same angle, the perpendicular to both meets at infinity, not at the centre of the turning circle. This forces the front tyres to slide sideways, causing an energy loss.

Steering geometry set-ups ensure that front wheels don’t turn the same angle.

tyre companies would love it if they did turn at the same angle.

Absolutely.

:)

Bogsnorkler said:

Michael V said:

mollwollfumble said:

Accelerator off. Traction control, either on or off.
Is there a big difference in final speed between turning while in gear and turning while in neutral?

My reason for asking is that conservation of energy suggests 60 km/hr but I’ve noticed while driving that it’s much closer to 0 km/hr. So steering can cause a massive energy loss, almost as much as braking. But why? I can’t seem to be able to draw force diagrams that make sense.

One thing that is clear is that because both front wheels turn the same angle, the perpendicular to both meets at infinity, not at the centre of the turning circle. This forces the front tyres to slide sideways, causing an energy loss.

Is this the dominant energy loss? Or is it the sliding sideways of both front and rear tyres under centrifugal action? How to calculate friction of the tyres on the road?

Physical testing in a real car?

One thing that is clear is that because both front wheels turn the same angle, the perpendicular to both meets at infinity, not at the centre of the turning circle. This forces the front tyres to slide sideways, causing an energy loss.

Steering geometry set-ups ensure that front wheels don’t turn the same angle.

tyre companies would love it if they did turn at the same angle.

Oh good. Thank you. That helps a lot. There’s brief mention on the last page of http://www.valleyofhastings.com/WheelAlignment2.pdf but I don’t fully understand the 3-D geometry.

> Physical testing in a real car?

I’ve learnt one thing. I’m nowhere near as brave as I thought I was.

In my car without traction control I don’t like driving full lock on suburban streets faster than 20 km/hr. That’s 30 km/hr into the turn, 20 km/hr out of the turn. At slower speeds the energy loss is less. 15 km/hr into the turn is about 15 km/hr out of the turn.

As for the physics, time to turn back to the “Physics of Motorsport” page http://community.dur.ac.uk/r.g.bower/public_html_old/PoM/pom/pom.html

From physics of motorsport link.

Maximum cornering speed v = sqrt (mu * r * g)

g = 9.81 m/s^2
r = 11 m (from initial question)
mu = 1.0 (for a road tyre)

Maximum speed v = 10.4 m/s = 37 km/hr

Well, the car is certainly not going to be able to go around that corner at 60 km/hr.

mollwollfumble said:

From physics of motorsport link.

Maximum cornering speed v = sqrt (mu * r * g)

g = 9.81 m/s^2
r = 11 m (from initial question)
mu = 1.0 (for a road tyre)

Maximum speed v = 10.4 m/s = 37 km/hr

Well, the car is certainly not going to be able to go around that corner at 60 km/hr.

Landy doesn’t slow down at all going around corners just turn on the inertia dampeners and can turn with any loss of speed

D.

To go that fast around a corner at that speed you’d need racing rubber and some downforce. Without those a regular road car will simply slide.

mollwollfumble said:

I’ve learnt one thing. I’m nowhere near as brave as I thought I was.

That’s one of the tricks you have to learn in motorsport – You have to be able to ignore the walls and trees and speed. It’s not easy to do.

FWIW the FSAE cars I’ve helped with could do 60 km/h around an 11 metre radius circle quite easily. With reasonable racing tyres they can generate about 1.5 G’s lateral with zero downforce, more as the speed builds up so the wings start to work.
My old racing car has big slicks and a bit of aero, it’ll generate about 2.2 G’s lateral at high speed and without a fair bit of practice that’s really hard on the neck muscles. You find that you simply can’t hold your head up after a few minutes.

mollwollfumble said:

If I’m driving a car at 60 km/hr and suddenly make a left 90 degree turn on hard lock, turning circle 11 metres.

How fast am I going at the end of the turn?

A) 60 km/hr (from conservation of energy)
B) 0 km/hr (from conservation of momentum)
C) Other speed
D) Not enough information
E) Can’t be done, the wheels will skid.

Spiny Norman said:

E.

To go that fast around a corner at that speed you’d need racing rubber and some downforce. Without those a regular road car will simply slide.

Bingo, thanks for sorting that out. And it agrees with the physics of motorsport website, above about 37 km/hr a typical (20 year old) road car will skid. You’d have to get a mu near 2.5 to get complete traction.

Spiny Norman said:

mollwollfumble said:

I’ve learnt one thing. I’m nowhere near as brave as I thought I was.

That’s one of the tricks you have to learn in motorsport – You have to be able to ignore the walls and trees and speed. It’s not easy to do.

FWIW the FSAE cars I’ve helped with could do 60 km/h around an 11 metre radius circle quite easily. With reasonable racing tyres they can generate about 1.5 G’s lateral with zero downforce, more as the speed builds up so the wings start to work.
My old racing car has big slicks and a bit of aero, it’ll generate about 2.2 G’s lateral at high speed and without a fair bit of practice that’s really hard on the neck muscles. You find that you simply can’t hold your head up after a few minutes.

Even 1 G laterally would scare me more than a little on a tight race track.

Which reminds me. My tablet has an inbuilt accelerometer, I keep meaning to try it out in the car. See what G forces I’m actually getting while I’m driving.

mollwollfumble said:

Which reminds me. My tablet has an inbuilt accelerometer, I keep meaning to try it out in the car. See what G forces I’m actually getting while I’m driving.

Yes, it does have an accelerometer, but can you get it to output its readings?

sibeen said:

mollwollfumble said:

Which reminds me. My tablet has an inbuilt accelerometer, I keep meaning to try it out in the car. See what G forces I’m actually getting while I’m driving.

Yes, it does have an accelerometer, but can you get it to output its readings?

I once had a science for kids app that will plot a graph of accelerometer readings. There’s probably something better out there.