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Old 8th July 2010, 16:10   #181
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Originally Posted by Samurai View Post
So that explains why your Jeep was struggling in some places.

For this discussion we have to consider healthy engine.

For a given slope, force of gravity on the Jeep will be same. I guess everybody can agree on that.

But the force exerted on the engine will differ based on the gears. It will be highest in lowest gear and gets less as we move higher in gear. This is due to decreasing multiplication factor in higher gears. That means Y rpm, where engine braking equalizes the force exerted by the wheels due to gravity will be highest at 1st gear, and gets lower at higher gears.

But there is a twist, the effect of higher momentum as the Jeep moves faster in higher gears. At this point mathematics has to take over and most of us are not capable of calculating all the forces involved here.

Samurai-san,

I think you got that backwards.

The force on the engine will be minimum in lowest gear and will rise subsequently with higher gears.
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Old 8th July 2010, 16:48   #182
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Originally Posted by ex670c View Post
You are right about the Reverse Gearing Effect, being Maximum, in 1st Low
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Originally Posted by '72 Bullet View Post
I think you got that backwards.

The force on the engine will be minimum in lowest gear and will rise subsequently with higher gears.
Ahem, you are saying it is absolutely opposite. I understand Arka's view, but can you explain why you feel otherwise?

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Originally Posted by ex670c View Post
You are right about the Reverse Gearing Effect, being Maximum, in 1st Low, but that is the point of equilibrium (Crawl Speed hence Torque), in 2nd Lo or after that the vehicle gains a lot of momentum and the reverse gearing starts acting
Does the point of equilibrium point remain the same in every gear? Won't the reverse effect be little higher in the next gear to compensate for the extra momentum? That would support 72Bullet's theory.
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Old 8th July 2010, 20:47   #183
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But Sutripta and Wolf specifically mentioned that simply going down the slope in gear will never lead to redlining. BTW, that is my observation too.

Why do you think they are wrong?

Since it is not hitting redline, then it is some rpm between idle and redline. If you don't want to call it Y, call it something else.
Hi,
I think I said
Quote:
Originally Posted by Sutripta View Post
Off hand, I'd say that the slopes necessary to redline an engine in the lower gears would cause you to break traction long before that.
I would not use the term "never" because I have a sneaking suspicion that there exists a class of vehicles in which it can be done.

I think that the question on equilibrium/ terminal speeds is a simple one, with a simple answer. And mostly answered without going into the intricacies of engines and gearboxes.

Bear with me while I bore you all with my thinking. Keep your fingers primed on the "gotcha" button.

Lets start with a simple question/ scenario:-
Suppose one has an infinitely long uniform slope. A car is allowed to coast down in neutral. No ABC. What will be the nature of time-speed graph? Will it have a terminal velocity? what will it depend on?

Regards
Sutripta

Last edited by Sutripta : 8th July 2010 at 21:07.
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Old 8th July 2010, 21:41   #184
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Originally Posted by Sutripta View Post
Lets start with a simple question/ scenario:-
Suppose one has an infinitely long uniform slope. A car is allowed to coast down in neutral. No ABC. What will be the nature of time-speed graph? Will it have a terminal velocity? What will it depend on?

Regards
Sutripta
Guys, your inputs please. Consensus needed before we can move on!

Sutripta
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Old 8th July 2010, 23:30   #185
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Coasting in neutral? How's that applicable to the whole engine braking concept?
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Old 8th July 2010, 23:39   #186
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^^^^
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Originally Posted by Sutripta View Post
I think that the question on equilibrium/ terminal speeds is a simple one, with a simple answer. And mostly answered without going into the intricacies of engines and gearboxes.

...

Lets start with a simple question/ scenario:-
Suppose one has an infinitely long uniform slope. A car is allowed to coast down in neutral. No ABC. What will be the nature of time-speed graph? Will it have a terminal velocity? what will it depend on?
Principles are the same. So we'll sort out the simpler stuff first, and then build it up.

Regards
Sutripta

Last edited by Sutripta : 8th July 2010 at 23:55. Reason: Reframed last line.
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Old 9th July 2010, 04:49   #187
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Quote:
Originally Posted by Samurai View Post

For a given slope, force of gravity on the Jeep will be same. I guess everybody can agree on that.

But the force exerted on the engine will differ based on the gears. It will be highest in lowest gear and gets less as we move higher in gear. This is due to decreasing multiplication factor in higher gears. That means Y rpm, where engine braking equalizes the force exerted by the wheels due to gravity will be highest at 1st gear, and gets lower at higher gears.

But there is a twist, the effect of higher momentum as the Jeep moves faster in higher gears. At this point mathematics has to take over and most of us are not capable of calculating all the forces involved here.
Sharath, The Force what you mentioned is converted into reverse torque. We call it as negative torque. Positive torque is when engine drives the wheels. Negative torque, when wheels drive the engine.
For Positive torque: the torque is calculated at the wheels.. Engine torque X Gear ratio X Diff ratio
For Negative torque: the torque is calculated at the engine.. Wheel torque X 1/Diff ratio X 1/Gear ratio. Now please re do the math.
And when wheels drive the engine, its a dead engine. If you want to drive the engine at Y rpm, you need X kw of power(product of rpm and torque). More rpm you want more amount of power you need to apply. Engine braking is measured in terms of KW or HP. The Exhaust brakes are the factors that multiply the engine braking power.. i.e xHP X FACTOR. where FACTOR can be 1 to x(x>1) depending on position of the exhaust flap. I hope this gives you some picture. I cannot be more clear on this.

When you down shift the gear you are reducing the power that is driving the engine. in which case the power you applied is not sufficient to keep up the rpm and will in turn retard.
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Old 9th July 2010, 11:13   #188
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[quote=Samurai;1973084]Ahem, you are saying it is absolutely opposite. I understand Arka's view, but can you explain why you feel otherwise?



Samurai-san,

A simplistic approach:

case 1 - Gear used - 4th H (1:1)
Going down a slope, suppose torque (force) experienced at wheel hub is 10 Nm. Assuming that the transmission losses are negligible, torque experienced at crankshaft/flywheel will be 10 Nm as well.

case 2 - Gear used - 1st L (40:1)
Going down the same slope and experiencing the same 10 Nm torque at wheel hub. Again assuming losses to be negligible, torque experienced at crankshaft/flywheel will be 10Nm/40 due to the reduction.

Therefore the torque/force experienced at the engine is less in lower gears and increases as you use higher gears.
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Old 9th July 2010, 11:33   #189
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Quote:
Originally Posted by Sutripta View Post
Hi,
I think I said

I would not use the term "never" because I have a sneaking suspicion that there exists a class of vehicles in which it can be done.

I think that the question on equilibrium/ terminal speeds is a simple one, with a simple answer. And mostly answered without going into the intricacies of engines and gearboxes.

Bear with me while I bore you all with my thinking. Keep your fingers primed on the "gotcha" button.

Lets start with a simple question/ scenario:-
Suppose one has an infinitely long uniform slope. A car is allowed to coast down in neutral. No ABC. What will be the nature of time-speed graph? Will it have a terminal velocity? what will it depend on?

Regards
Sutripta

Regarding the time-speed graph, I expect it would start of with a gradual slope and slowly get steeper till it finally hits a plateau (terminal velocity). This is assuming time is on horizontal axis and speed is on vertical axis.

On an unlimited slope, terminal velocity will be reached when
acceleration = drag
here acceleration will be due to gravity and mass of the vehicle.
Drag will be due to aerodynamic drag, rolling resistance of the wheels and other rotating parts (axles, diffs, propeller shaft, gears in the t.case/g.box)
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Old 9th July 2010, 13:15   #190
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Default Terminal Velocity

Quote:
Originally Posted by Sutripta View Post
Hi,
Lets start with a simple question/ scenario:-
Suppose one has an infinitely long uniform slope. A car is allowed to coast down in neutral. No ABC. What will be the nature of time-speed graph? Will it have a terminal velocity? what will it depend on?
Hi Sutripta,

I think there are a few Factors which will affect the Terminal Velocity.

1) Inclination of the Slope.

2) Resistance of Drive line components, Co-Efficient of Friction (Tyre & Surface), COG of Vehicle, Aerodynamics

The Vehicle will gain Terminal Velocity, when the momentum initiated by Gravity overcome the Resistance, which is holding the vehicle.

Regards,

Arka
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Old 9th July 2010, 20:42   #191
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Guys, thanks for participating. Good to see that we are on the same page regarding the vehicle reaching terminal velocity!

Quote:
Originally Posted by '72 Bullet View Post
Regarding the time-speed graph, I expect it would start of with a gradual slope and slowly get steeper till it finally hits a plateau (terminal velocity). This is assuming time is on horizontal axis and speed is on vertical axis.
Less steep with increasing time, actually.

On an unlimited slope, terminal velocity will be reached when
acceleration = drag
Lets call it the propulsive force. Drag is of course the retarding force.

here acceleration will be due to gravity and mass of the vehicle.
Drag will be due to aerodynamic drag, rolling resistance of the wheels and other rotating parts (axles, diffs, propeller shaft, gears in the t.case/g.box)

Quote:
Originally Posted by ex670c View Post
Hi Sutripta,

I think there are a few Factors which will affect the Terminal Velocity.

1) Inclination of the Slope.
Agreed. Higher gradient = higher terminal velocity

2) Resistance of Drive line components, Co-Efficient of Friction (Tyre & Surface), COG of Vehicle, Aerodynamics
Why CoG? Also, think pt 2 best be called rolling resistance of tyres.

The Vehicle will gain Terminal Velocity, when the momentum initiated by Gravity overcome the Resistance, which is holding the vehicle.
Think 72 Bullet's explanation is better worded.

Regards,

Arka
There are two points to note
a) if the propulsive force is constant (as it is for an uniform slope) and the resistive force is an increasing function of velocity (known by all TBhpians), even without knowing the actual numbers, we can state with certainty that there will be a terminal velocity. (Given the numbers, or actual graphs, we can state what that terminal velocity will be)

b) We have reached our conclusion without having to decompose/ resolve our resistive forces (into say this is for transmission loss, and that is for aerodynamic drag etc). It is because we are seeing the vehicle as a system.

Let us now add some resistive force, and work out its effect on the terminal speed.

Assume in our idealised world, brakes are perfect and do not fade!
We apply the handbrake till the shoes just engage. (case 2)
And then we apply it another click. (case 3)

How will our three (no handbrake, handbrake just applied, handbrake applied a bit more) speed-time graphs vary.
I think it can be answered easily, intutively, and correctly.

Let us now add another variable. Can we increase the slope so that the terminal velocity in case 3 is more than the terminal velocity in case 1 with the original slope?

Regards
Sutripta

Last edited by Sutripta : 9th July 2010 at 20:51.
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Old 11th July 2010, 10:49   #192
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Assume in our idealised world, brakes are perfect and do not fade!
We apply the handbrake till the shoes just engage. (case 2)
And then we apply it another click. (case 3)

How will our three (no handbrake, handbrake just applied, handbrake applied a bit more) speed-time graphs vary.
I think it can be answered easily, intutively, and correctly.

Let us now add another variable. Can we increase the slope so that the terminal velocity in case 3 is more than the terminal velocity in case 1 with the original slope?

Regards
Sutripta[/quote]

case 2 vs case 1
I guess the graph will be similar in shape. Only difference is that as the resistive force is actually greater in case 2, it will accelerate at a slower rate. Also, terminal velocity will be less. It will be the similar for case 3, only slower.

In theory, it is possible to increase the gradient till the terminal velocity at case 3 will be higher than terminal velocity of case 1 (which is on a slope less steep).

Regarding the original time-speed graph, I think the graph will get steeper as it progresses, as in the beginning it takes a little bit of time to build up speed after which the acceleration builds up faster. This is from my perception of rolling down a slope / a ball rolling down a slope.
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Old 11th July 2010, 16:33   #193
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^^^
Agree with you. TrmVel(1) > TrmVel(2) >TrmVel(3)
Also, increasing the slope (ie propulsive force) will increase the Terminal velocity.
Don't agree with you regarding the velocity graph becoming steeper with time. Maybe we can debate it later.

The important thing is that the system (vehicle) does not care about the source of the additional retarding force. It can be brakes, or rolling resistance, or a strong head wind, or regenerative braking, or engine braking. It is just easier to comprehend when it is brakes. The result is the same:-
Increased resistive force => decreased terminal velocity .

An engine being motored will need power. To spin it faster, one needs more power. And once again the engine doesn't care how it is being motored. Its power dissipation characteristics will remain the same.

For any given road speed, the engine will spin faster in a lower gear.

Putting all these together, one gets that coming down a slope in any gear with throttle closed, some terminal velocity will be reached, and that lower the gear, lower the terminal velocity. But there is no magic number to this velocity. It will vary depending on a number of parameters.

Regarding the question of redlining the engine during engine braking.
As I said, I have a gut feel (ie can be totally wrong).
A modern car is supposed to start up a 1 in 3 slope fully laden. I feel that such a car will not redline during engine braking.
However we have a large class of vehicles designed with a single goal:- maximum load- minimum fuel. These tend to be the grossly underpowered people or goods movers. One sees them all the time, really struggling to move up slopes which we wouldn't even notice in normal cars. I feel there is a high chance that these vehicles, given the right (or is it wrong) set of circumstances, can redline their engines during engine braking.

E&OE

Regards
Sutripta

Last edited by Sutripta : 11th July 2010 at 16:48.
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