Physical relationships of Speed, Drag, Force, and Power (split from: Begode ET)

[Robse]

3 minutes ago, InfiniteWheelie said:

No, because of wind resistance. 2x the speed requires 8x the power. The free spin could be 300 km/h, but you still won't have enough power to go much over 100 km/h.

4x the power, not 8.     e=1/2 mv^2

[InfiniteWheelie]

5 minutes ago, Robse said:

4x the power, not 8.     e=1/2 mv^2

I'm pretty sure that's not correct. I think it's 8x the power to go double the speed, but only 4x the energy per km since you're going twice the distance per second.

Edited January 2 by InfiniteWheelie

[Robse]

13 minutes ago, InfiniteWheelie said:

I'm pretty sure that's not correct. I think it's 8x the power to go double the speed, but only 4x the energy per km since you're going twice the distance per second.

Distance lies within the speed.  In the ideal universe the kinetic energy is equal to the energy used to archive the speed given.  1/2mv^2 dictates that the kinetic energy of an object with the mass X will be 4 times higher if speed is doubled. Same with braking distances versus speed. Its 4 times.  But nevertheless 4 or 8, its pretty much more energy you need to travel at 160 kmph versus 80

[RagingGrandpa]

24 minutes ago, Robse said:

e=1/2 mv^2

That's the kinetic energy of the moving body... but it has little to do with the energy per second (power) lost to drag, which is the power the EUC needs to supply.
 

30 minutes ago, InfiniteWheelie said:

2x the speed requires 8x the power.

Regarding aero drag, agreed.

[Robse]

4 minutes ago, RagingGrandpa said:

That's the kinetic energy of the moving body... but it has little to do with the energy per second (power) lost to drag, which is the power the EUC needs to supply.
 

Regarding aero drag, agreed.

this is getting complicated, since wind drag is exponential, and the continued need for power can be even more than 8x at higher speeds.

[RagingGrandpa]

For a given drag constant K...

V1 = 10 (m/s)
F1 = K V1 V1 = 100 K
P1 = F1 V1 = 1000 K

V2 = 20
F2 = 400 K
P2 = 8000 K

P2 = 8 P1

2x the speed requires 8x the mechanical power.
 

Edited January 2 by RagingGrandpa

[Robse]

4 minutes ago, RagingGrandpa said:

V1 = 10 (m/s)
F1 = K V1 V1 = 100 K
P1 = F1 V1 = 1000 K

V2 = 20
F2 = 400 K
P2 = 8000 K

P2 = 8 P1

2x the speed requires 8x the mechanical power.

it looks like you have v1 ^3 ? why? 🙂

[RagingGrandpa]

Drag Force = K V²

Power = Force * Speed

P = K V³

[Robse]

10 minutes ago, RagingGrandpa said:

Drag Force = K V²

Power = Force * Speed

P = K V³

ok, but rhen explain this: my wife has a ebike, a speed pedelec, goes 45 kmph (legal) the motor is 680 watt. Its has a advanced display in direct connection with the hardware. When se rides at 20 kmph the power consumption is 160 watt.  When se rides at 40 kmph the consumption is 590 watt. that's not even 4 times the 20 kmph consumtion ?

[RagingGrandpa]

13 minutes ago, Robse said:

When she rides at 20 kmph the power consumption is 160 watt.
When she rides at 40 kmph the consumption is 590 watt.

There are more drag forces than Air involved.

The "rolling resistance" (from the tire and machine) scales more slowly than V². So at low speeds, rolling resistance dominates. At high speed, aero drag dominates.

If there were a 20kph tail wind, I think she would still need some 100 watts to sustain that 20kph.
And the additional ~50W is needed for aero drag, if there were no tail wind.

So perhaps ~200W rolling resistance @ 40kph.
And some 400W for air drag. That's 8x drag.
 

Edited January 2 by RagingGrandpa

[Robse]

hahaha  ...  i was scrolling up and down, wondering where all the other comments about the ET MAX where.....

[techyiam]

2 hours ago, Robse said:

4x the power, not 8.     e=1/2 mv^2

Energy = 1/2 mass x velocity^2

This computes the kinetic energy a mass has while travelling at a specific speed.

 

Aerodynamic drag force opposes motion.

That means a vehicle need to generate a force to overcome the drag force.

Energy = Force x Distance.      ( 1 joule = 1 newton meter )

This is the energy required to generate a force by a vehicle to overcome the aerodynamic drag force acting on it over a specific distance travelled.

Power = Force x Velocity.        ( 1 watt = 1 newton meter per second )

This gives the power required to generate the force by a vehicle to overcome aerodynamic drag for the vehicle travelling at a specific speed.

Edited January 2 by techyiam

[Robse]

1 hour ago, InfiniteWheelie said:

I'm pretty sure that's not correct. I think it's 8x the power to go double the speed, but only 4x the energy per km since you're going twice the distance per second.

Looks like you are right, and i'm not

[InfiniteWheelie]

This is actually particularly interesting for electric wheels.

As you approach top speed on any type of vehicle, you lose acceleration the closer to top speed you get, which isn't a problem. However this is a problem for us because if we lean (attempt to accelerate) too hard when close to top speed, we cut out. This means we have to not only watch our top speed, but also our acceleration when nearing it.

The cool thing is because of the exponential nature of wind resistance, we can keep massive power buffers without missing out on much top speed. For example let's say you have a 13kw wheel, and it takes 10kw to go 100km/h (roughly true). This mean's at 100 km/h you have 30% reserve power (70% PWM). HOWEVER if you used all 13kw, you'd only gain 10 km/h.

Say we made a 20kw wheel by using high power cells/motor/controller (possible today). This leaves us 50% reserve power at 100 km/h, but you give up only around 25 km/h. With enough power you get a wheel that's very fast, has huge reserve power, AND is nearly impossible to overpower (regardless of lean, hills, headwinds etc), without giving up much speed. The faster the wheel, the more useless it is to chase the (increasingly less) unused speed at the expense of safety margin. The future is ultra powerful wheels with massive power reserves, which give up very little top speed to achieve.

Edited January 3 by InfiniteWheelie

[InfiniteWheelie]

Here's a table to help people see the speed/power relationship for wind resistance. The power numbers aren't exact but they're pretty close (the ratios are correct though).

Edited January 3 by InfiniteWheelie

[Jason McNeil]

I'll defer to @RagingGrandpa superior wisdom, man this guy knows everything , but the drag coefficient cross-section of a standing Rider has like twice the amount of drag than the squatting position, this is without doubt the greatest factor for reducing energy consumption at speed.  
https://www.researchgate.net/publication/321866543_Calculation_of_Aerodynamic_Drag_of_Human_Being_in_Various_Positions.
Approximate values of the drag coefficient (c w ), according to [8], are obtained experimentally and shown in figure 10: standing person 0.78, cyclist in an upright position 0.53-0.69, cyclist in bent position ~0.4.

[techyiam]

1 hour ago, Jason McNeil said:

but the drag coefficient cross-section of a standing Rider has like twice the amount of drag than the squatting position, this is without doubt the greatest factor for reducing energy consumption at speed.  
https://www.researchgate.net/publication/321866543_Calculation_of_Aerodynamic_Drag_of_Human_Being_in_Various_Positions.

This paper documents CFD simulation modelling a wind speed of equivalent to 38 km/h.

CFD is a computational model solving the Full Navier Stokes equations. This is a set of partial differential equations:

(1) Equation of Motion.

(2) Conservation of Mass

(3) Consevation of Momentum

(4) Conservation of Energy

(5) Conservation of Mass (aka Continuity). 

They used a simplified human model, and assume steady flow.

But they did include a turbulence model.

But the results weren't verified in a wind tunnel. 

For people who have ridden at speeds standing and sitting would probably have a better feel for aerodynamic drag.

Edited January 3 by techyiam

[Mono]

2 hours ago, Jason McNeil said:

Approximate values of the drag coefficient (c w ), according to [8], are obtained experimentally and shown in figure 10: standing person 0.78, cyclist in an upright position 0.53-0.69, cyclist in bent position ~0.4.

Keep in mind that the drag force is proportional to the area of attack and to the drag coefficient and squatting reduces (obviously) the area of attack by a rather significant margin too! The figure shows the much less obvious reduction in the drag coefficient. The combination of both, area and coefficient, may reduce the overall drag force due to squatting+bending by a factor of four-or-so (twice a factor of two-or-so).

https://en.wikipedia.org/wiki/Drag_equation gives the drag force formula.

Edited January 3 by Mono

[techyiam]

5 minutes ago, Mono said:

Keep in mind that the drag force is proportional to the area of attack and to the drag coefficient and squatting reduces (obviously) the area of attack may a rather significant margin too! The figure shows the much less obvious reduction in the drag coefficient.

It would be way to complex to figure Cd numerically.

Reducing frontal area is certainly the most obvious way to reduce aerodynamic drag force.

There is nothing aerodynamic about a dude riding an euc down the highway at 60+ mph. There are no aerodynamics aids.

One can experiment with where to place the arms, maybe.

Without a rider on an euc in a wind tunnel, it would be difficult to figure out something meaningful.

[Mono]

1 hour ago, techyiam said:

There is nothing aerodynamic about a dude riding an euc down the highway at 60+ mph. There are no aerodynamics aids.

The aerodynamics aid is changing the shape from standing to squatting+bending. As the figure and @Jason McNeils comment suggest, squatting+bending are aerodynamically advantageous (by quite a significant amount) compared to standing upright, because the figure gives the Cd for different body positions (the Cd, not the drag force). That's actually not really a big surprise for anyone somewhat familiar with Cd values.

Edited January 3 by Mono

[techyiam]

3 minutes ago, Mono said:

The aerodynamics aid is changing the shape from standing to squatting+bending. As the figure and @Jason McNeils comment suggest, squatting+bending are aerodynamically advantageous (by quite a significant amount) compared to standing upright, because the figure gives the Cd for different body positions (not the drag force). That's actually not really a big surprise for anyone familiar with Cd values.

There are two positions riders have tried on euc's, and that is standing and sitting. 

And we already know sitting gives less drag.

Anyone riding at high speeds can tried the two positions and would know right away which position would give less drag.

[Mono]

On 1/3/2024 at 3:05 PM, techyiam said:

It would be way to complex to figure Cd numerically.

Complex, yes, too complex, I don't know. I could be wrong, but I think that current state-of-art drag simulation models will do a pretty good job to approximate such Cd values.

On 1/3/2024 at 4:43 PM, techyiam said:

There are two positions riders have tried on euc's, and that is standing and sitting. 

And there is bending while standing and there is sitting upright versus bending down while sitting. What we don't know, for example, is whether bending while standing produces more or less drag force than sitting upright. The latter has most probably a larger Cd but may have a smaller front area. I'd probably bet on bent standing to have less drag force if I needed to bet.

On 1/3/2024 at 4:43 PM, techyiam said:

And we already know sitting gives less drag.

Right. That was not the point of any of my previous comments though which related to Jason's post on how much the drag coefficient is modified by different positions.

Edited January 5 by Mono

[Robse]

seated riding gives me +20% more range

[Mono]

2 minutes ago, Robse said:

seated riding gives me +20% more range

Without knowing the speed, this is a meaningless number, as there exists for sure for any rider and (almost) any wheel a speed where this is actually true.

[Robse]

1 minute ago, Mono said:

Without knowing the speed, this is a meaningless number, as there exists for sure for any rider and (almost) any wheel a speed where this is actually true.

of course you can not use my 20% in a formula, it's not the same from rider to rider, different wheels, speed, road condition etc  etc. However, for me, that number is nearly the same on two different wheels, and once i had the opportunity to try another wheel for 3 days, it was again about the same 20%.  But just to verify: seated riding reduces wind drag "a lot" 😉