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#### colinh

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Hi all,

having read the "I want to build an EV! Where do I start?" sticky thread - here I am The idea would be for the following

very light (500 kg) - like a Caterham (weight and torque-wise) but a completely different (highly aerodynamic) shape. Welded steel space frame, with aluminium panels.

high peak torque (500 Nm) - EMRAX 268 synchronous PM motor (20 kg)

"direct drive" (ie. no transmission (gear box). Just a rear LSD (limited slip differential).

medium range (16 kWh battery) - (approx 100 kg) 200 km ?

high performance (t 0-100 kmh = 5 s, Vmax = 160 kmh) - OK, 160 kmh isn't really high speed for a "high performance" vehicle, but in practice that's my comfortable cruising speed on an Autobahn, and if you want to get any sort of range at all, you have to go pretty slow (100 kmh) anyway.

My (back of envelope) calculations so far:

The EMRAX 268 MV has a max peak torque (T.pk.mot) of 500 Nm and max continuous rotation speed (w.max.mot) of 4500 rpm. If LSD is 3:1 that gives peak torque (T.pk.whl) of (500 x 3) 1500 Nm and max speed (w.max.whl) of (4500 / 3) 1500 rpm at the wheel.

"Normal" wheels (225/55 R16 ... 235/35 R19) have radius (r.whl) 0.32 m and circumference (C.whl) 2.01 m

Fmax = T.pk.whl / r.whl = 1500 Nm / 0.32 m = 4688 N

Vmax = w.max.whl / 60 * C.whl = 1500 rpm / 60 s/min * 2.01 m = 50.25 m/s (= 180.9 kmh)

acc.max = F.max / m = 4688 N / 600 kg = 7.8 m/s2 (=> 28 m/s / 7.8 m/s2 = t.0-100 = 3.6 s)

----

Power is required to overcome drag (wind resistance) and rolling resistance (tyres).

F.drag = 1/2 * rho * v^2 * [ c.d * A]
: 0.5 * 1.22 * 0.25 * v^2
: 0.15 v^2

where rho is air density = 1.22 kg/m^3
c.d is drag coefficient
A is frontal area.
A very aerodynamic car might have a [c.d * A] of 0.25 m^2.

F.rr = c.rr * F.n = 0.01 * 750 kg * 9.8 = 74 N

where c.rr (coefficient of rolling resistance) is ~ 0.01
F.n is normal (downward) force which is 750 kg * 9.8 m/s^2

F.drag is proportional to v squared. F.rr is more or less constant.

Energy is force x distance. Power is Energy per second = force * distance per second = force * speed.

P = (0.15 * v^2 + 74) * v.

[ 160 kmh = 160,000 / 3600 = 44.4 m/s (/2.01 * 60 = 1320 rpm (wheel) = 3960 rpm (mot) ]

P.160 = ( 0.15 * 44.4^2 + 74) * 44.4 = 370 N * 44.4 m/s = 16.5 kW

"Free running losses" in the motor are given as ~ 1.9 kW at 4000 rpm.

P.tot.160 = 16.5 + 1.9 = 18.4 kW

[ 70 mph = 112 kmh = 31.3 m/s = 934 rpm.w = 2800 rpm.m]

P.112 = ( 0.15 * 31.3^2 + 74) * 31.3 = 221 * 31.3 = 6.9 kW
P.frl = ~ 1 kW
P.tot.112 = 7.9 kW

16 kWh battery would give 2 hrs at 70 mph. ie 200 km range.

The EMRAX 268 MV "medium" voltage requires up to 680 V and can cope with up to 400 A (for max torque, when accelerating for a few seconds).

----

I haven't really looked into the battery yet. Using NCR18650B cells as a nominal unit (3.6 V, 3.35 Ah [= 12 Wh] 2C = 6.7 A, 48.5 g, 5 EUR):

16 kWh = 1333 cells. = 65 kg.
400 A (peak, only) requires 60 cells in parallel (at 2C, continuous)
180 cells in series = 650 V.

Clearly I don't want 180 x 60 = 10,800 cells = 130 kWh = 525 kg = 54,000 EUR.

If I have 4 groups of 45 x 15 that would be 2700 cells, 32 kWh, 130 kg and 13,000 EUR.

These could be switched in series to give 650 V and 100 A, or parallel to give 162 V and 400 A or 2x2 giving 200 A at 325 V.

I'm surprised that I can't find larger/better batteries than 18650s! 15 in parallel would give me a 3.6 V, 50 Ah [180 Wh] cell at 0.73 kg and 75 EUR.

A GWL 3.2 V, 20 Ah LiFePO4 prismatic cell is 64 Wh at 0.65 kg and 25 EUR.

200 => 12.8 kWh, 130 kg, 5000 EUR.

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