Dropping to flat

supersonic

Ages since I studied physics!

I am pretty sure the land at speed phenomenon is due to the way we lift the front end when going much slower and the weight we transfer.

yoohoo999

TommyEss wrote:

yoohoo999 wrote:

i'm a corporate lawyer

Boo... Hiss... And other assorted pantomime noises!

ha ha, nothing like a bit of prejudice! I actually act in the interests of the general public. I deal with the legals on the 2012 Olympics, hospitals, schools etc.

i actually get to help people which is cool

GHill

yoohoo999 wrote:

As I understand it, the bike would hit the ground with the same kinetic energy regardless of whether the rider dropped to flat or hit it at 100mph, provided that his y axis speed (ie the drop speed) was the same.

Therefore, if the kinetic energy is identical, surely all the difference in speed, or indeed technique in landing does is dissipate that energy by a different (and hopefully more efficient) manner. Down to the action of the suspension, rider position, rider movement etc.

Or am I missing something?

Yes, you are missing something. The bike isn't just being dropped vertically (y axis in the above frame of reference), you also have a horizontal component that means you are hitting the ground at some sort of diagonal (angle will depend on many factors). So, as mentioned above, you need to combine the two components.

yoohoo999

why is my mass hitting the ground at an angle? my mass is travelling down the y axis, reaching a final velocity of 5.4m/s, my mass obviously hasn't changed, therefore why would this affect the kinetic energy calculation below?

KE = 1/2 (M * (V * V))

TommyEss

If you came off the ledge at speed rather than just dropping off the edge, then your mass has a momentum in a diagonal downwards direction - downwards due to the mass's acceleration due to gravity, and "across" i.e. away from the ledge in the direction of travel, due to velocity in the direction you were riding.

supersonic

I'm tired lol, so forgive me if I get confused with terms, but once resolved, the force in the Y direction is only dependant on the starting height.

yoohoo999

tommy, i understand that, but how would that momentum alter the kinetic energy value as I hit the ground?

if someone can set out a formula which demonstrates that the kinetic energy of an object is greater/less upon impact due to its x axis velocity then you will have solved the issue at the heart of this entire thread.

I can't find one and for that reason I think the kinetic energy value stays the same.

supersonic is right. there are 2 movements, but the kinetic energy on impact is only concerned with one of them, the y axis (height).

albo

You take Vectors in X and Y axis to be separate, the Horizontal component wouldn't effect your Vertical ( hitting the floor )

GHill

supersonic wrote:

I'm tired lol, so forgive me if I get confused with terms, but once resolved, the force in the Y direction is only dependant on the starting height.

If only want the force along the y axis, but who cares about that? It is not equal to the overall force or energy of the bike as it hits the ground.

TommyEss

Ah - but energy can neither be created nor destroyed, only transferred from one type to another.

In that way, you approach a ledge with some kinetic energy, and some gravitational potential energy.

You then ride off the ledge, and all of that gravitational potential is turned into the following:

Kinetic - the speed of you vertical descent

Sound - you'll make a thud as you land

Thermal through Friction/Wind resistance - a tiny amount, granted...

The forward kinetic energy you have will not really be converted to anything other than a bit of thermal friction against the air - so if you ride off at 15mph you'll land and continue forwards at pretty much 15mph (assuming you land it well)

The thing is that they're independent of each other - if you hit the jump at twice the speed, you'll travel (pretty much) twice as far - and when you land your forwards velocity (velocity is a vector so has a value and a direction) will be twice as high - but your vertical velocity - the "speed" at which you hit the ground, will be the same)

If that doesn't make sense, you really need some nice line diagrams with arrows showing how the vector splits.

Imagine a boat crossing a stream. It has propulsion pushing it from one bank to the other, acting perpendicular to the direction of the flow of the river.

The river has a flow of water at right angles to the boat's desired direction of travel.

If the boat and the water are flowing at the same SPEED, then the velocity of the boat travelling over the water, will be at 45 degrees - it will go, for example, one metre closer to the bank for every one metre it's pushed downstream by the river.

albo

But only Vertical momentum will be transferred, Horizontal will be conserved almost entirely ( depending on certain small factors )

TommyEss

Yep - because you have two vectors, you have kinetic energy in the vertical plane (the GPE converted) AND kinetic energy in the horizontal plane (the initial ride forwards, minus not a lot)

yoohoo999

GHill wrote:

supersonic wrote:

I'm tired lol, so forgive me if I get confused with terms, but once resolved, the force in the Y direction is only dependant on the starting height.

If only want the force along the y axis, but who cares about that? It is not equal to the overall force or energy of the bike as it hits the ground.

eh? of course the height of the drop dictates the overall force of the bike hitting the ground! are you honestly trying to say that a bike dropped from 100m will hit the ground at with the same force as a bike dropped from 1m? that's just nuts.

the simple fact as I see it, is that your decelleration as you hit the ground is IDENTICAL when you land on flat from speed or just dropping off. If you looked at a graph you would hit the ground in the exact same time flying off the end as you would just dropping off (the whole bullet dropped/bullet fired analagy works here).

therefore the force of hitting the ground is the same. the reason 45 degree transitions feel smoother is because you decellarate slower.

albo

Energy is not a vector, you need to be talking about Momentum

supersonic

It depends on what we are looking at and how we are modelling it.

Force transmitted into the ground (lets assume a hard surface and point mass) is the same regardless of X velocity. If the ground and air was frictionless, the X component could be infinite - the drop would feel the same when you landed!

However the ground is not like this, is not hard, does have friction and the decceleration due to this will cause a mass transfer on landing.

GHill

If you want to convince yourself that the horizontal motion makes a difference think of a car dropping off a small curb. At 5 mph the effect will be a bump but nothing to worry about. At 150 mph it will be a big deal (serious or possibly terminal damage).

Don't try this at home.

supersonic

Bloody hell, you lot type fast! My last post is six post old already!

GHill

yoohoo999 wrote:

GHill wrote:

supersonic wrote:

I'm tired lol, so forgive me if I get confused with terms, but once resolved, the force in the Y direction is only dependant on the starting height.

If only want the force along the y axis, but who cares about that? It is not equal to the overall force or energy of the bike as it hits the ground.

eh? of course the height of the drop dictates the overall force of the bike hitting the ground! are you honestly trying to say that a bike dropped from 100m will hit the ground at with the same force as a bike dropped from 1m? that's just nuts.

That's not what I said at all. I'm simply saying that you can't separate the variables in this case.

TommyEss

Yeah - sorry - the terms have been getting confused...

Speed is not the same as velocity

Energy is not the same as force

Mass is not the same as momentum

So what's your real question?

yoohoo999

TommyEss wrote:

The thing is that they're independent of each other - if you hit the jump at twice the speed, you'll travel (pretty much) twice as far - and when you land your forwards velocity (velocity is a vector so has a value and a direction) will be twice as high - but your vertical velocity - the "speed" at which you hit the ground, will be the same)

so on that basis, surely the actual impact felt by the rider will be identical regardless of the speed you hit the jump at, since the only conversion of energy that will be felt will be that of the gravitational kinetic energy?

albo

GHill, I can see exactly where you come from, before I studied Physics I saw it the same, everybody in my class did!
But you have to appreciate that components can be split up into X and Y, and don't effect each other. But, I do appreciate where you are thinking from.

The reason the bike will retard is because the downwards force increases the friction between the tires and ground, almost no other reason is valid... I think

yoohoo999

TommyEss wrote:

So what's your real question?

to quote myself, this is the overall question of the thread I think:

so on that basis, surely the actual impact felt by the rider will be identical regardless of the speed you hit the jump at, since the only conversion of energy that will be felt will be that of the gravitational kinetic energy?

albo

As long as it is a flat-to-flat drop, yes

supersonic

yoohoo999 wrote:

TommyEss wrote:

So what's your real question?

to quote myself, this is the overall question of the thread I think:

so on that basis, surely the actual impact felt by the rider will be identical regardless of the speed you hit the jump at, since the only conversion of energy that will be felt will be that of the gravitational kinetic energy?

For a smooth hard surface, yes ;-)

TommyEss

But the impact of landing throws the rider in a certain direction too - as he is falling with the bike, so the rider has a vertical and horizontal component too.

As Supersonic said - the friction generated on landing on soft ground plays a big role in the way the rider is kicked about on landing - if you land on soft ground, then there is a massive negative acceleration retarding against the direction of forward movement, as well as the "thud" downwards landing component.

If you are going faster when you hit the jump, you may find that instead of digging in, you skim a bit more on landing - this would greatly reduce the decrease in forward movement so you'd find the landing a lot lighter.

albo

So we can conclude that:

The Impact felt by a rider A on a bike B dropping from a flat take-off to a smooth hard flat landing will be constant.

GHill

albo wrote:

GHill, I can see exactly where you come from, before I studied Physics I saw it the same, everybody in my class did!
But you have to appreciate that components can be split up into X and Y, and don't effect each other. But, I do appreciate where you are thinking from.

Can you show me an equation for why this is true (sorry that I won't just accept your word ). How can an object hitting something diagonally not be affected by its horizontal motion? Are you telling me that a bullet dropped from 1' onto polystyrene will embed itself just as far as a bullet fired horizontally from the same height? (given a very long sheet of polystyrene)

TommyEss

Dropped from one foot up the bullet will have gravitational potential energy equal to it's mass in kg multiplied by the acceleration due to gravity squared.

When fired from a gun I think you'll find it has rather more energy imparted to it from the gunpowder than a one foot drop would give it, so it's not a fair comparison.

albo

Imagine throwing a ball into the air 10m and letting it hit your head, and then (somehow) dropping a ball 10m onto your head the impact felt will be identical.

Does that analogy help?

Bullets spin, and that is how they manage to stay so straight, and stop from falling so fast. I don't like thinking about them

yoohoo999

Very fun and interesting thread, i can't believe I started all this with "is my rear wheel dropping to much?"