# Creating a Realistic Car Simulation Game

• crazyone
In summary: It will be different depending on the tire's size and the grip level of the surface on which the car is driving.In summary, the speed of the car is related to the speed of the tire, but the wheel circumference cannot always be calculated using just the circumference of the tire. The wheel circumference is also affected by other forces, such as friction.
crazyone
Hi,

I'm making a car simulation game not just a simple one, one that closely matches the kinds of Gran Turismo, therefore, it has to be extremely precise and realistic and use real life physics.

I have been searching around for information regarding this subject and was able to get some info here and there about parts, about maths but it's hard to put it all together correctly and that's where you guys could help me out.

The first thing I'm trying to establish and confirm is the following question:

Speed of an object such as a car is directly related to the rotationnal speed of the tires right? If a tire is 15in radius, thus he is grossly 94.25in circumference and if i roll the wheel once, the car should then move by 94.25 inches right?

If this is true, then i can assume that a car's engine turning at 3000 RPM or 50 Main shaft revolutions by second, will spin the wheel by 50 turns each seconds. In between that comes the transmission which may drastically lower this ratio to make the car pickup inertia more easily.

So if i have a transmission that has a coeficient of 4.00 in the first gear, the calculation should be the following if i am not wrong:

(These numbers are grossly calculated to remove the numerous decimals but will be kept in the final calculations.)

Engine Revs / s = 50
Gear 1 ratio = 4.00
Wheel Revs / s = (50 / 4) = 12.5
Distance covered / s = (94.25 * 12.5) = 1178 inches (or 98 feet)
Speed = 66 mph

If i put that in miles per hour it makes the car roll at 66 miles per hour... This makes no sense right? I mean, a car in the fifth gear could do that speed around 3k RPM, not at gear 1. Same thing again, with gear 5 having a ratio of 0.98:

Engine Revs / s = 50
Gear 1 ratio = 0.98
Wheel Revs / s = (50 / 0.98) = 51
Distance covered / s = (94.25 * 51) = 4806 inches (or 400 feet)
Speed = 272 mph

What am i not grasping correctly in your opinion?

Your gear ratios are tall. In a Corvette Z06, which can reach 61mph at 7000 rpm in first gear, first gear has a ratio of 2.66 to 1, while the differential gearing is 3.42 to 1, for an overall gearing of 9.10 to 1. The rear tire size is 325/30/19, the specificication include an overall diameter of 26.7 inches, and 783 revolutions per mile.

At higher speeds, the driven tires have to opposed rolling resistance and aerodynamic drag, and the tire tread deforms before and during contact with the pavement and then recovers aftewards. This reduces the effective diameter by a small amount at higher speeds, maybe 1% or so.

Hey thanks for the reply but that was not it but it did raise some new questions. The original problem was the wheel circumference calculation, i thought the R## was the radius of the wheel which is wrong. I can tell you how to calculate it but i think it is beyond the scope of this post.

My new question that arises is the following, what is a differential gear? You are saying that at one point, what i am aware of so far in the car movement mechanics is the following:

Engine -> Revs
Revs -> Attenuation from clutch pressure plate / Torque converter attenuation
Converter -> Transmission Gearing
Transmission -> Wheels

Where does the differential go in there and what is it?

ps: Indeed the gear ratios where high, i was trying to match the speed of a toyota corolla at certain RPM and found that the only way was to achieve this by tweaking the transmission gear ratios... But now that the wheel circumference is good, i can put much more logical and reallistic numbers in the gear ratios and everything falls into place.

crazyone said:
Speed of an object such as a car is directly related to the rotationnal speed of the tires right? If a tire is 15in radius, thus he is grossly 94.25in circumference and if i roll the wheel once, the car should then move by 94.25 inches right?

Well.. yes.. but that is true considering pure rolling. In that case, it can be said that:

$$v_cm = r\omega$$

As you said, you want to use real-life physics.. in that case, pure-rolling is something that never happens. However, all objects try to attain the state of pure rolling.

When working with a car, one major force you need to account for is friction. That and other forces combined, you can get to the fact that for a wheel, the motion will be a combination of the translation motion of the wheel and rotational motion of the wheel. So, you can say that:

$$v_cm = v + r\omega$$

Do note that, $v_cm$ will definitely be equal to the translational velocity of the car. This equation becomes even more complex when you consider the car under acceleration. The frictional constant, isn't a constant in true sense. The frictional constant is a function of the speed and shape of the acting body. This function can however be approximated to a simple mathematical function.. something that may not be true.. but for a sim.. will be adequate.

Yeah that i was aware of, i knew that there are all kinds of factors that may affect the distance, what i'd like to point out though, right away, i am not a mathematician. Just a more than average programmer that knows what he's doing, so usage of VcM = RW is like pure mystery for me.

Please use less mathematical formulas and more explanations :) I like your description of pure-roling though, i can understand that.

So, if i undestand your post correctly, friction can affect the speed of the car? If so how? Is it calculatable?

crazyone said:
So, if i undestand your post correctly, friction can affect the speed of the car? If so how? Is it calculatable?

it definitely does.. Frictional force is calculated using:

$$f_s = \mu_s N$$

here, 'N' is the normal force i.e. the force exerted by the object [in our case the car on the surface]. The variable $\mu_s$ is a characteristic of the pair of surfaces in contact and is different for different surfaces. As i said earlier, this is considered to be a constant but well.. in the case of physics of fast moving objects like a car, it changes with change in the speed and shape.

Look up all that goes into translating engine rpm into wheel rpm, you missed the differential in your example, which drops speed by its divisor effect. If the engine is turning at a certain rpm, then the car WILL be going at the speed indicated by the net gearing.

The question of whether the engine WILL turn at that rpm is a function of the power output versus the power required to move the car at the speed suggested by the net gearing, which is primarily moving air out of the way.

Ok, so i'd need to evaluate the friction characteristic for asphalt which would give me the "Ug" sign (Not sure if it's a Ug, don't even know how you are displaying this). Can you tell me where to find such an information?

Also, what is N, the weight of the vehicule in this case? I would also need to take into account the pressure of the tires and the shock absorber's value i guess?

Thanks regor60, don't know if i updated about the speed yet so here goes what i have now.

Name Corolla
Weight 2850,00
Inertia 0,00
RPM Increase Max 0,00
Base Torque 162
At Base RPM 4000
RPM 6000
Torque 243
Power /s 24300
Shaft rev / s 100,00
Torque converter applied friction 100%
Torqueconverter conversion ratio 100%
Flywheel conversion ratio 100%
Transmission rev / s in 100,00
Active gear 5
Gear 1 Ratio 316%
Gear 2 Ratio 190%
Gear 3 Ratio 1310%
Gear 4 Ratio 88%
Gear 5 Ratio 72%
Gear 6 Ratio 0%
Transmission rev / s out 138,89
Differential ratio 265%
Revolutions after differential ratio 52,41
Applied brake force 0%
Revolutions after braking 52,41
Tire Quality P
Tire width (mm) 215,00
Tire W/h Ratio (%) 45%
wheel size length cm 62,53
m/s 32,77
m/h 117981,13
km/h 117,98
mile/h 73,74

As you can see, to match the speed of a corolla at 3000 RPM (i own one and i do about 120 km/h) i have to put the engine RPM at 6000 following all the formulas from the excel sheet i have. So something is definately missing here, i have a 0.5:1 factor missing somewhere in between.

What do you think that could be in terms of mechanics?

Oh and please, let's just keep the conversion about car speed calculation for now, thanks for the help rohanprabu, but i'd like to focus only on one aspect at a time.

When the speed calculation works, i'll focus on calculating the accelleration potential...

It's not clear from your list of data what the tire diameter is. If it's 63 cm, the speed at 6000 rpm in 5th gear would be (52 rev/s) (3.14*.63 m/rev) = 100 m/s. Of course, the engine could never get the car going that fast. Although most cars will not reach redline in top gear, this gearing looks too far off to be right. Searching around a bit, I think your axle ratio is wrong.

As a general warning, the level of physics in something like Gran Turismo is at a far higher level than the calculations we're discussing. If you're still interested in this sort of thing, there's a lot of documentation over at racer.nl. Someone there has already done what you're proposing.

thanks, i'll look it up, note that i want to make a racing sim but it's an online one, so I'm not looking at doing an extreme racing sim, just one that is not a online, click and flip a coin in hopes to win, i want some really mechanical aspects and allow my players to tune the cars.

I'll go check the site you are talking about

I've looked it up, it's not showing much other than presenting the racing engine, i found a part discussing car physics, but so far i know as much as is presented.

After reading your post, I've tried to find out what you mean by 52 rev/s, i can't seem to produce this, where do you get that from?

6000 RPM = 100 Rev/s
After gear 5 at 82% (or 0.82:1), i get 121.95 revolutions,
After differential at 345% (or 3.45:1) i get 35.35 revolutions this is far from the 52 you are stating...

(PS i may have changed the numbers trying to find data on the net, so you don't have the same numbers as me anymore)

Finaly, yes the tire size is simple to calculate, the formula is:

(Wheel Radius (15) * 2.54 (to get CM)) + ((185mm * 65%)/10) and this gives us ... O_O

Oh ****, i got it, this is stupid, i am not having the circumference here, I'm simply calculating the tire's height, i need to multiply it by PI.

OMG, YES! That was the missing part, that's why i had crappy km/h. Now were talking... The speed for my car at 3000 RPM is now 124km/h which is much more logical.

Thanks

Ok, now the second part which i have NO CLUE or almost what to do, power calculations. Now that i know how fast i can go i need to calculate how a car can accelerate based on it's power output.

crazyone said:
I've looked it up, it's not showing much other than presenting the racing engine, i found a part discussing car physics, but so far i know as much as is presented.

This seems to have some more detailed discussion:

http://racer.nl/prgusers.htm

There are (or were) descriptions of the Pacejka model for the tires somewhere too. I think some older versions of the software are actually open source, which may be useful to you.

Ok, now the second part which i have NO CLUE or almost what to do, power calculations. Now that i know how fast i can go i need to calculate how a car can accelerate based on it's power output.

If the wheels aren't spinning, a reasonable first approximation to the acceleration is found from
$$m a v = \epsilon P_{\mathrm{eng}} - P_{\mathrm{drag}}$$
m is mass. a is acceleration. v is velocity. $\epsilon$ is a driveline efficiency (near 1). $P_{\mathrm{eng}}$ is the engine power at the appropriate rpm. The last term represents power lost to aerodynamic drag and rolling resistance of the tires. It can be approximated by
$$P_{\mathrm{drag}} = \alpha v+ \beta v^3$$
for appropriate constants $\alpha$ and $\beta$. These terms can be estimated by searching around for tire (first term) and aerodynamic (second term) data.

I know, that a car has Torque@RPM, that produces torque in a matter of foot-pounds or SI units. I know i have first to calculate how much power the engine can output at the RPM it's reving. What i am not sure is if the RPM of the engine affects the torque at the same time.

For example, if a car can output 162ft-lbs at 4000 rpm, the total power output, as I've read about it, is 648000 SI per minute or in this case 10800 SI per second. Now do we have to do a cross calculation based on the RPM to find the torque or is the torque always the same at different RPM? My guess is that the torque changes with the RPM but then that would mean that the multiplicator has a double edge, you would get very low SI when the engine is idling.

700 rpm * 162 / 4000 = 28.35 SI * 700 RPM / 60 seconds = 330.7 SI... This is almost 3% of the optimal possible SI capable by the engine.

Can anyone confirm that?

Thanks for the info stingray, the data presented in that page is about the engine itself, i did peak at the car physics at the bottom but this is way to mathematical for me, i need to understand first then i can maybe start playing around with formulas.

I'm not a mathematician, just a simple programmer that is maybe more advanturous than others :)

Ok if i understand your post stingray, the movement of a car is equivalent to it's mass multiplied by acceleration multiplied by velocity and this must equivalent to the SI units generated by the engine and i'd have to remove the drag from this.

Taking into account that i don't have any information relative to drag i will simply assume it is 0 for now and try to understand what you are telling me.

If the velocity of my car is 0, because we are not moving and the mass is 2850 pounds, and the SI units generated from my engine are 330 units, i can calculate the acceleration by doing the following:

Mass * Accel * Velocity = SI
Mass * Accel = SI / Velocity
Accel = SI / (Mass * Velocity)

Problem is, with this formula, if velocity is at 0, this creates a division by 0 error. Note that my math courses are quite far away, is this the right way to alter a formula to isolate a new result?

Another thing I'm wondering, the engine will still be able to rev up but slowly at the first calculation pass creating more power on the second pass and so on. So if i have a strong enough power, i can eventually move right? That would be how a car could eventually defeat hills and wind drag?

Yeah, the equation I gave doesn't easily work for zero speed. This is related to the fact that an engine developing any power must act through a slipping driveline (clutch or torque converter) in order to start a car moving. The effective efficiency goes to zero for a moment. Alternatively, you might want to think about an electric motor. This works even without rotation, but it generates no mechanical power at the moment that nothing's moving. It is generating a torque (like an internal combustion engine acting through a slipping clutch), and that can be used to find the initial acceleration. You can work with torque instead of power the whole way through, but it's more complicated. You then have to worry a lot more about gear ratios.

In answer to your other question, max torque does depend on rpm. Look up dyno curves to get an idea of how things work. In your game, the torque (or power) function can be approximated by a polynomial.

Back to your specific torque calculations, I get 162 ft-lb = 220 N-m (Newton-meters are the SI unit of torque). The power at this rpm is then (220*4,000/60) = 150,000 W (Watts are the SI unit of power). If the torque remained at its peak value all the way down to idle, the power available there would be down by a factor of (4,000/700) = 5.7. In reality, the torque will be much less at idle. The idle power will therefore be much worse than this.

Tire diameter = wheel diameter + 2 x (tire width x aspect ratio). In your case

Tire diameter = 17 inches + 2 x (215mm x .45) = 17 inches + 7.62 inches = 24.62 inches.

If your car has an automatic transmission, then you get a torque multiplier effect from the fluid clutch. If you car has a manual transmission, some of the engine's energy is converted to heat when the clutch is slipping. The fluid cluth also consumes energy throug conversion to heat. Some automatics have a "lock up" clutch, essentially a manual clutch that doesn't slip once at speed and at lower throttle pressure.

The maximum torque output from the engine will vary with rpm.

When launching from a standing start, if the engine is allowed to spin at higher rpms, then the effective rear wheel torque is limited to the dynamic friction at the clutch and the friction of the tires (static if they aren't slipping, dynamic if they are). Most street cars have clutches that don't grip well if there's a big difference between engine rpm's and output shaft rpms.

Assuming a manual transmission, once the clutch is no longer slipping, then the rear wheel torque will equal the engine torque multiplied by the overall gear ratio, divided by the effective radius of the rear tire, and reduced by the losses in the drive train. Drive train losses average about 15%, depending on the car.

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Thanks to both of you, ok, so the SI units are not the same as Torque ft-lbs? I'll look up a converter on the net and create a conversion function.

Stingray, if the equation you provided doesn't work for 0 speed, then what will? There must be a way to extrapolate the required force needed to move the car when velocity is 0 no? Then if you have the minimal power you could extrapolate from the max power the engine is capable of delivering the movement that is applied to the car's movement no?

For torque nominal function, i'll look it up, i guess it must be some kind of inversed U graphic type function that i can't remember the name and the peak power@rpm is the top of the U graphic. That i can eventually build myself if i can't find it.

Jeff, thanks for confirming the tire and other info. Do you know where i could find info relative to transmission ratios, clutch friction reduction aspects and torque converter reduction effects (This must be a function since they work with moving fluid compared to a clutch which is often a disk of metal or ceramic)

Good evening to all

crazyone said:
So the SI units are not the same as English ft-lbs?
No.

Force
1 Newton = 0.224809 lb

Distance
1 meter = 3.28084 ft

Mass
1 kg = 0.068521766 slug
1 slug = 32.174 lb (mass)

Power:
1 horsepower = 550 lb ft / sec.
1 watt = 1 Newton meter / second = 0.001341022 hp

Pacejka model ... doesn't work for 0 speed, then what will?
Pacejka model is a crude model used to simplify tire grip calculations. It works backwards, using speed and slippage as inputs to determine forces. It fails to work at zero speed and also at very slow speeds.

Obviously it's not a problem for a car to take off from a standstill. Torque from the engine is multiplied through the transmission and differential to provide a torque at the rear wheel. From a standing start, the clutch allows the engine to spin at a different speed than the ouput shaft of the clutch that goes into the transmission, with some loss of power due to conversion to heat. Using my previous example where a drivetrain losses about 15% of the power, just use a guestimate that combined with slipping clutch, the losses are about 20%. So rear wheel torque will equal 80% of engine torque times the overall gear factor. If the tires don't slide, then the torque divided by the radius of the driven tires determines the force applied to the pavement, which in turn provides a reactive force that is equal to the force from the tires, and provides the actual accelerating force on the car. If the tires slide, then the engine torque is spent increasing the angular speed of the tires as well as accelerating the car forward.

Using a simple case, assume that the tires are not slipping, and that the engine is only using partial throttle to produce 200 ft lbs of torque (rpm doesn't matter since clutch is slipping). Using my "guestimate", slipping clutch and drivetrain losses reduces this to 160 ft lbs of torque, but then this torque is multiplied by the overall gearing in 1st gear, 8.374 in your case, So there is 1339.84 ft lbs of torque applied to the rear tires. Tire radius is 24.62 inches, or 2.0517 feet so rear wheel force is 1339.84 ft lbs / 2.0512 ft = 653.198lbs. The car weighs 2850 lbs, so it's mass is 88.581 slugs. The acceleration is 653.198lbs / 88.581 slugs = 7.374 ft / sec^2, or 0.2292 g's.

Note that some of this torque is being used to accelerate all the moving parts in the drivetrain, trasmission gears, drive shaft, differential, rear axle (or half shafts), rear wheels and rear tires. I don't know how much this loss is, maybe 2% of the torque depending on which gear the car was in (less for taller gears), and this may be included in the 15% loss factor I often see quoted.

Unlike the Pacejka model, for straight line acceleration, I believe that rear wheel torque forces should be calculated first. If the tires are not spinning, then acceleration corresponds to rear wheel torque directly. If the tires are spining, then the acceleration of the car is a function of the dynamic friction of the tires times their normal force, and the remainder of the torque is being used to accelerate the drivetrain as mentioned above, except the losses would be more than 2%.

To make things even more complicated, the dynamic friction of the tires decreases as the difference in speed between tire surface and pavement increases. This varies depending on the tire.

There's a guy named Todd Wasson who has been working with car simulation physics:

http://www.performancesimulations.com

A company named iRacing will be releasing a racing game soon, and they have collected a large amount of real world data, and claim that their soon to be released racing game will be the most realistic:

http://www.iracing.com

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Jeff Reid said:
Pacejka model is a crude model used to simplify tire grip calculations. It works backwards, using speed and slippage as inputs to determine forces. It fails to work at zero speed and also at very slow speeds.

I only mentioned the Pacejka model in passing. The things I was explaining had nothing to do with it. Still, I think it is a reasonable thing to use in a simulation. Slip angles and ratios need to be calculated to have any semblence of realistic behavior. I don't know what else you can use for that without oversimplifying things. The problem is getting good data for all the input parameters. I don't like the model myself, but just about everything publically available seems like a kludge.

Stingray said:
I only mentioned the Pacejka model in passing. The things I was explaining had nothing to do with it. Still, I think it is a reasonable thing to use in a simulation. Slip angles and ratios need to be calculated to have any semblence of realistic behavior. I don't know what else you can use for that without oversimplifying things. The problem is getting good data for all the input parameters. I don't like the model myself, but just about everything publically available seems like a kludge.
In my opinion, once a large amount of good data is obtained, a better approach is to use very large tables of data, with linear or polynomial interpolation, with multiple tables to handle the hysteresis effect of how cars behave when at or past the limits and recovery once past the limit. The Pacejka model is good, but not great, and falls apart in very simple situations such as a car at rest on a banked track. It doesn't deal with the hysteresis effects that occur when transitioning just above and below the limits. Most games with Pacejka model revert to a spring like model for low speeds.

A much better model would be one based on actual tire data, longitudinal and lateral forces versus slip ratio versus slip angle versus normal force, including the hysteresis effect that the resultant grip force difffers depending if these factors are approached from above or below. After all, this is how cars work in real life. The starting point is engine, braking, and steering inputs, plus track angle, track and tire friction characteristics, and the final result is the longitudinal and lateral forces. Pacejka model starts from the middle, slip ratio and slip angle, and tries to work it's way outwards, a simple but not all that accurate approach.

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God and i thought i was intelligent :P

Thanks guys, i will re-re-re-read the post from jeff about the accelelation calculation and try to apply it to the data i currently have. If it works, then i will use this to implement accelleration.

Later guys

Thanks Jeff, now i understand your instructions better, the only thing i don't see in there is how to apply current velocity/inertia of the car to acceleration. I was thinking of the previous model the Pacejka, could i simply apply it using a 0.01 velocity value or simply skip it in the formula if the velocity is at 0? I don't see a problem to that since it's the first run, after that the car is running so i can imply the velocity into the formula.

Your formula is great i just don't see where velocity comes into play.

crazyone said:
Thanks Jeff, now i understand your instructions better, the only thing i don't see in there is how to apply current velocity/inertia of the car to acceleration. I was thinking of the previous model the Pacejka, could i simply apply it using a 0.01 velocity value or simply skip it in the formula if the velocity is at 0? I don't see a problem to that since it's the first run, after that the car is running so i can imply the velocity into the formula.

Your formula is great i just don't see where velocity comes into play.

You should probably ignore our little tangent on the Pacejka model. It actually has nothing to do with the equation I wrote down. That only uses the definition of mechanical power as the rate at which work is being performed. For nonzero speeds, it's actually identical to the procedure Jeff outlined. In symbols, his statement was essentially that (ignoring efficiency issues)
$$m a = g \tau_{\mathrm{eng}}/r ,$$
where g is the overall (transmission and differential) gear ratio, r is the tire radius, and $\tau_{\mathrm{eng}}$ the torque exerted by the engine. You can convert to the equation I first gave by using the fact that power and torque are related through the engine's speed. If the clutch or torque converter isn't slipping,
$$P_{\mathrm{eng}} = \tau_{\mathrm{eng}} gv/r .$$

Velocity doens't affect real wheel torque, but does affect rolling resistance and aerodymamic drag. Assuming that nothing is slipping (clutch or tires), then rear wheel rpm = engine rpm divided by gear ratio, and rear wheel torque = engine torque at that engine rpm times the gear ratio, minus the drive train losses (the 15% I mentioned before is close enough), and Stingrays equations describe the math properly.

Stingrays equations work fine once the clutch is not slipping. Unless you have real world data, you're going to have to guestimate as to how much power is converted into heat while slipping the clutch, combined with the fact that friction in street clutches is reduced as the angular speed difference between input and output plates increases, and if the speed differential is high, then maxium torque from the clutch is less than maximum torque the engine can produce. For a racing car or aftermarket clutches, the clutches use materials that will end up handling all the torque an engine can produce even when there's a lot of slippage.

What you'll need an alternative method for is how to implement grip forces when tires are near the limits of traction (both below and above). The other issue is obtaining real world data for tire characteristics, and this varies based on tire construction and compounds used. Pacejka tries to simplify this, and it's a close approximation, but fails to take into account hysteresis which has a significant effect on real cars, and it's really not all that accurate when right at the limits.

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## 1. How do you accurately simulate the physics of a car in a game?

To create a realistic car simulation game, extensive research and understanding of physics is necessary. The game engine must accurately calculate and simulate the forces acting on the car, such as gravity, friction, and air resistance. Developers also use real-world data and formulas to ensure the car's movement and handling are as close to reality as possible.

## 2. What factors are considered when designing the graphics for a car simulation game?

The graphics in a car simulation game play a crucial role in creating a realistic experience. Developers consider factors such as lighting, textures, and shading to accurately depict the car's appearance in different environments. They also pay attention to details like reflections, shadows, and particle effects to enhance the overall visual quality of the game.

## 3. How do you create realistic audio for a car simulation game?

Sound plays a significant role in creating a realistic car simulation game. Developers use recordings of real cars to create accurate engine sounds, tire screeches, and other vehicle noises. They also take into account factors like the car's speed, terrain, and weather conditions to ensure the audio matches the gameplay.

## 4. What challenges do developers face when creating a realistic car simulation game?

Developing a realistic car simulation game can be a challenging task. One of the main challenges is balancing realism with gameplay. While it's essential to accurately simulate the physics and mechanics of a car, the game must also be enjoyable for players. Developers must also optimize the game to run smoothly on various devices and platforms.

## 5. How do you incorporate different car models and brands into a simulation game?

Car simulation games often feature a variety of car models and brands for players to choose from. To incorporate these vehicles into the game, developers must obtain licenses and permissions from the manufacturers. They also need to gather data and specifications for each car to accurately simulate its performance and handling in the game.

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