What are the powertrain calculations for the E-Baja competition?

It is similar to m-BAJA powertrain calculation.

First of all, you should find drag forces on the vehicle then find out the torque on wheels, after that decide your reduction on the basis of your motor output torque. Its depend on you that which type of transmission system you choose (auto or manual). Then find your acceleration, top speed and gradeability of vehicle.

For the basics take help from fundamental of vehicle dynamics by Thomas d. Gillespie and for coefficients and constants take help from google search engine.

Manual Transmission calculations – 2 | Tractive forces, accelaration and gradability

Some basic formula for the calucation of various dynamics of vehicle…..for gradability the force component will be resolve in two component one is along the road and one is perpendicular to the road. only along the road component will help in climbing and perpendicular to the road will cause the vehicle to topple. there are various losses in engine , tranmission due to various friction so the result which is gonna come is gonna be in ideal case(i.e no friction etc) so what ever you get after calculation the actual result will be only 60-70 % of that. so calculate accordingly.
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here is sample calculation(  ideal case)

At hill climbing, tractive force available at an inclination of 45°,
F=Wsinθ- μWcosθ ≈2081N
Power =10hp=7460 Watt =F*V
V=3.58m/sec=12.9km/hr
So, our vehicle can climb the hill of inclination 45° with this velocity.
On a horizontal road,
Air resistance (Ra)= Ka*A*(V)²
 Ka= 0.004725 – 0.005 ,depending upon type of vehicle we have considered it to be 0.005
 Area (A) = Projected front area= 1.025 m²
 Velocity = 59.06km/hr
Hence Ra=17.88N
Rolling Resistance (Rr) = K*W
 W=total weight of vehicle= 300kg
 K is a constant and depends upon the nature of surface
 We have considered the road surface to be loose mud whose rolling resistance is 0.0435
Hence Rolling resistance (Rr) = 128.02N
Total resistance= Ra + Rr =145.9N
Traction –
F= (Te* Gear ratio* Transmission efficiency)/wheel radius
 Te=engine torque= 19Nm
 Wheel radius = 0.3175m
Hence F=756.21N (at low speed)
Acceleration:
a=(F-R)/m=2.03m/sec²
So, 0-60 km/hr in 8.21 sec.

 

Manual Transmission Gearbox – How it Works!

“Automatic vs manual” is a debate that has gone on for years and isn’t coming to a stop any time soon. Even though we don’t see that debate ending, we think that it might be beneficial to drop a little knowledge that could aid in the situation.

Today, we show you what exactly makes a manual transmission tick. What exactly happens when you bang those gears to get your car or truck rolling?

From why the transmission is even necessary to how it takes the power created from the engine and transfers it to the wheels, this demonstration breaks it all down.

Check out the video below to learn a few things about how all of this works. Better break out that notepad and write a few things down because we’re going back to school.

 

CALCULATIONS FOR FNR GEAR BOX AND CVT COMBINATION

Download : CALCULATIONS FOR FNR GEAR BOX AND CVT COMBINATION

Gear Reduction Selection

Power Requirements against:

 

Fr: Wheel Resistance

FL : Air resistance

Fs: Gradient resistance

Fa: Acceleration resistance

 

Wheel Resistance= Rolling resistance + Slip Resistance + Road Resistance

Tr: Brake Torque

 

Assumption taken

 

Unmade road coefficient of friction fr= .16

mass of vehicle being=Mf

 

CALCULATIONS

 

Fr=fr*Mf*g*Cosαst     where αst=30˚

 

Fr=0.16*200*9.81*Cos 30

Fr=280N

 

Total Load R=113kg(driver)+200kg (vehicle)

R=313*9.81=3070.50N

 

Static Resistance=µR

=0.65*3070.5=1995.84N

 

Vmax=60kmph

Wheel Radius r=221.875mm

= .221m

Air Resistance

 

FL:0.5ρCwAv2

Ρ=Air Density

Cw=Drag Coefficient

A=Area

V=velocity

 

FL =0.5*1.99*0.0315*(16.66)2

=0.52N

 

 

Gradient Resistance:

Fst=Mf*g *Sinαst

 

Putting the corresponding value

Fst=131.725N

 

Acceleration resistance:

Fa=m*a

=956N

Ftotal=2662.247

 

Power Required: Ftotal*v/(Nrpm*η)