Abstract
Braking at half is easily taken to halve the margin for turning. What this paper shows is that tyre friction is shared between braking and turning within one circle, so a little braking barely reduces the turning force while hard braking reduces it greatly, and that stopping while turning takes $\sin^{-1}y/y$ times the distance of a straight stop ($y$ being the fraction of friction used sideways). No new theorem or law is claimed.
First, the fraction of friction left for turning is $\sqrt{1-b^2}$——using a fraction $b$ for braking leaves $\mathbf{0.866025}$ at $b=0.5$ (addition would say $0.5$) and $\mathbf{0.435890}$ at $b=0.9$ (Section 2).
Second, the first braking barely touches the turning grip——using $30$ per cent of the friction for braking reduces the turning grip by only $\mathbf{4.6061}$ per cent; using $90$ per cent reduces it by $\mathbf{56.4110}$ per cent (Section 2).
Third, the smallest radius one can turn grows $\mathbf{1.154701}$-fold with half braking and $\mathbf{2.294157}$-fold with nine-tenths——at $60$ km/h with friction coefficient $0.8$, $35.4068$ m becomes $40.8843$ m and $81.2288$ m (Section 3).
Fourth, and this is the core. Stopping while turning takes $\sin^{-1}y/y$ times the distance of a straight stop——taking a $50$ m radius curve at $60$ km/h and braking with all the friction left over from turning, the car stops in $\mathbf{19.6714}$ m, $\mathbf{1.111163}$ times the $\mathbf{17.7034}$ m of a straight stop; the closed form and numerical integration agree (Section 4).
Fifth, near the limit the ratio approaches $\pi/2$——for sideways use $y$ of $0.5$, $0.9$ and $0.99$ the ratio is $\mathbf{1.047198}$, $\mathbf{1.244188}$ and $\mathbf{1.443694}$, tending to $\pi/2=\mathbf{1.570796}$ as $y\to1$; exactly at the limit, all the friction is used sideways and braking cannot begin (Section 4).
Sixth, the separator is whether braking and turning friction are treated as adding like numbers or as a quantity with direction——the friction limit applies to the combined length (Section 5).
Keywords: friction circle, tyre, braking, cornering, vehicle dynamics.
Scope (what is not claimed)
No new theorem or law is claimed——the friction circle, centripetal acceleration and stopping distance are all standard. The car is a point mass and the friction circle is an idealisation——a real tyre has different limits along and across (an ellipse), varying with slip angle; here it is a circle. A friction coefficient of $0.8$ is representative of a dry road. Load transfer (weight moving to the front wheels under braking), anti-lock systems, tyre elasticity and road banking are not included. No driving advice is given——this paper does not recommend any particular way of driving. Relation to earlier papers: Paper 453 treated a banked curve on an icy road as a point mass and found the band of safe speeds——it used only the sideways friction. This paper adds friction along the road (braking) to the same point mass and shows that the limit applies to the combined length. Paper 300 showed that sharing a root is decidable——stopping on a straight and stopping on a curve come from the same friction circle and share a root; only the sideways use $y$ separates them ($y\to0$ returns the straight). What is added is setting the fraction left for turning beside the additive assumption, deriving the closed form $(R/2)\sin^{-1}(v^2/(R\mu g))$ for stopping while turning and checking it against numerical integration, showing that the ratio $\sin^{-1}y/y$ tends to $\pi/2$ at the limit, and placing the separator on addition versus a quantity with direction.
1. Notation and the Question
Setting: treat the car as a point mass and split its acceleration along the road into a longitudinal part (braking) $a_x$ and a lateral part (turning) $a_y$. The limit of acceleration the tyres can supply, $\mu g$, applies to the combined length:
$$\sqrt{a_x^2+a_y^2}\le\mu g .$$Using a fraction $b=a_x/(\mu g)$ for braking leaves $\mu g\sqrt{1-b^2}$ for turning.
Observation 1.1 braking at half is easily taken to halve the margin for turning. Friction is seen as one purse paid out to braking and to turning——whether the purse empties by addition or in some other way is another question.
Question: how much does using friction for braking reduce the turning force, and how far does it take to stop while turning?
2. Circle and Addition
Computation 2.1: the fraction $b$ used for braking and the fraction left for turning.
| $b$ | circle $\sqrt{1-b^2}$ | loss of turning grip | additive assumption $1-b$ |
|---|---|---|---|
| $0.10$ | $0.994987$ | $0.5013$ per cent | $0.900000$ |
| $0.30$ | $0.953939$ | $\mathbf{4.6061}$ per cent | $0.700000$ |
| $0.50$ | $\mathbf{0.866025}$ | $13.3975$ per cent | $0.500000$ |
| $0.70$ | $0.714143$ | $28.5857$ per cent | $0.300000$ |
| $0.90$ | $\mathbf{0.435890}$ | $\mathbf{56.4110}$ per cent | $0.100000$ |
| $0.99$ | $0.141067$ | $85.8933$ per cent | $0.010000$ |
Observation 2.2 the first braking barely touches the turning grip. At the top of the circle the fraction left falls only as the square of $b$: using $30$ per cent for braking costs only $4.6061$ per cent of the turning grip——the last tenth is the dearest; at nine-tenths, more than half the turning grip is gone.
3. The Radius One Can Turn
Computation 3.1: at $60$ km/h with friction coefficient $0.8$, the smallest radius $v^2/(\mu g\sqrt{1-b^2})$ is $35.4068$ m without braking, $40.8843$ m ($\mathbf{1.154701}$ times) with half the friction used for braking, and $81.2288$ m ($\mathbf{2.294157}$ times) with nine-tenths.
Observation 3.2 with half braking the radius widens by only $15$ per cent. Addition would double the radius; the circle gives $1.154701$——but at nine-tenths braking it is $2.294157$, and almost no margin for turning remains.
4. Stopping While Turning
Computation 4.1: a car on a curve of radius $R$ at speed $v_0$ brakes with all the friction left over from turning. At speed $v$ turning needs $v^2/R$, leaving a deceleration $\sqrt{(\mu g)^2-v^4/R^2}$, which grows as the car slows. The stopping distance is
$$s=\int_0^{v_0}\frac{v\,dv}{\sqrt{(\mu g)^2-v^4/R^2}}=\frac{R}{2}\sin^{-1}\frac{v_0^2}{R\mu g}, \qquad \frac{s}{v_0^2/(2\mu g)}=\frac{\sin^{-1}y}{y},\quad y=\frac{v_0^2}{R\mu g},$$with $y$ the fraction of friction used sideways.
| sideways use $y$ | speed | straight stop | stop on curve | ratio $\sin^{-1}y/y$ |
|---|---|---|---|---|
| $0.5$ | $50.4171$ km/h | $12.5000$ m | $13.0900$ m | $\mathbf{1.047198}$ |
| $0.708136$ | $60$ km/h | $\mathbf{17.7034}$ m | $\mathbf{19.6714}$ m | $\mathbf{1.111163}$ |
| $0.9$ | $67.6416$ km/h | $22.5000$ m | $27.9942$ m | $\mathbf{1.244188}$ |
| $0.99$ | $70.9431$ km/h | $24.7500$ m | $35.7314$ m | $\mathbf{1.443694}$ |
The radius is $50$ m. Numerical integration and the closed form agree to $5.0\times10^{-12}$ or better in all four.
Observation 4.2 this is the core. Stopping while taking a $50$ m curve at $60$ km/h takes $19.6714$ m, $1.111163$ times a straight stop——turning uses $0.708136$ of the friction sideways, so at first only $\sqrt{1-y^2}=0.706076$ is left for braking; as the car slows, turning needs less and more is left for braking.
Observation 4.3 near the limit the ratio approaches $\pi/2$. The limiting speed for the curve, $\sqrt{R\mu g}$, is $71.3005$ km/h; a straight stop from it takes $25.0000$ m, while on the curve the limit is $\pi R/4=39.2699$ m, a ratio of $\pi/2=\mathbf{1.570796}$——exactly at the limit, however, all the friction is used sideways and braking cannot begin; $\pi/2$ is where the ratio heads when the limit is approached from below.
5. The Separator
Observation 5.1 the separator is whether braking and turning friction are treated as adding like numbers or as a quantity with direction. The friction limit applies to the combined length, so the fraction left for turning is $\sqrt{1-b^2}$, not $1-b$——a little braking is cheap and hard braking is dear. That stopping while turning takes $\sin^{-1}y/y$ times as far comes from the same circle.
Observation 5.2 beside Paper 453. It treated a banked curve on ice as a point mass and found the band of safe speeds with sideways friction alone——this paper adds longitudinal friction to the same point mass. While friction is used for turning, there is that much less for stopping.
Observation 5.3 one root, under Paper 300. The straight and the curved stopping distance come from the same friction circle——only the sideways use $y$ separates them, and as $y\to0$ the ratio returns to $1$.
Observation 5.4 what is not said. No driving advice is given. Real tyres have different limits along and across, load transfers and anti-lock systems act, and none of these is included——what is shown is the form, a limit on a quantity with direction, and the closed form that follows.
6. Roll Call and Closing
- half braking leaves $\mathbf{0.866025}$ for turning (Computation 2.1): addition says $0.5$.
- nine-tenths braking leaves $\mathbf{0.435890}$ (Computation 2.1): $\mathbf{56.4110}$ per cent of the turning grip gone.
- $30$ per cent braking costs $\mathbf{4.6061}$ per cent (Observation 2.2): the first braking is cheap.
- the turning radius grows $\mathbf{1.154701}$- and $\mathbf{2.294157}$-fold (Computation 3.1): at half and nine-tenths.
- stopping while turning takes $(R/2)\sin^{-1}(v^2/(R\mu g))$ (Computation 4.1): a closed form.
- $\mathbf{19.6714}$ m on a $50$ m curve at $60$ km/h (Observation 4.2): $\mathbf{1.111163}$ times a straight stop; the core.
- the ratio is $\sin^{-1}y/y$ (Computation 4.1): $\mathbf{1.047198}$, $\mathbf{1.244188}$, $\mathbf{1.443694}$.
- near the limit, $\pi/2=\mathbf{1.570796}$ (Observation 4.3): exactly at the limit, braking cannot begin.
- integration and closed form agree to $5.0\times10^{-12}$ (Computation 4.1): two routes.
- the separator is addition versus direction (Observation 5.1): the limit applies to the combined length.
Closing: Braking at half is easily taken to halve the margin for turning. Tyre friction shares one circle between braking and turning——using half the friction for braking still leaves $\mathbf{0.866025}$ for turning, and $30$ per cent braking costs only $\mathbf{4.6061}$ per cent of the turning grip; but nine-tenths leaves only $\mathbf{0.435890}$. At $60$ km/h with friction coefficient $0.8$, the smallest radius one can turn grows $\mathbf{1.154701}$-fold with half braking and $\mathbf{2.294157}$-fold with nine-tenths. Stopping while turning takes $\sin^{-1}y/y$ times the distance of a straight stop——on a $50$ m curve at $60$ km/h the car stops in $\mathbf{19.6714}$ m, $\mathbf{1.111163}$ times the $\mathbf{17.7034}$ m of a straight stop. As the sideways use $y$ approaches $1$ the ratio approaches $\pi/2=\mathbf{1.570796}$; exactly at the limit, all the friction is used sideways and braking cannot begin. The separator is whether braking and turning friction add like numbers or form a quantity with direction——the limit applies to the combined length; under Paper 300 the straight and curved stopping distances come from one circle and share a root. Placed among the earlier papers——Paper 453 found the band of safe speeds with sideways friction alone; this paper adds longitudinal friction to the same point mass. To be honest——the car is a point mass, the friction circle is idealised, and load transfer, anti-lock systems, tyre elasticity and banking are not included. The friction coefficient is representative, and this is not driving advice.
References
- Gillespie, T. D. (1992). Fundamentals of Vehicle Dynamics. Society of Automotive Engineers.
- Milliken, W. F., & Milliken, D. L. (1995). Race Car Vehicle Dynamics. SAE International.
- Pacejka, H. B. (2012). Tire and Vehicle Dynamics (3rd ed.). Butterworth-Heinemann.
- Yamagishi, Y. (2026). The Three Hundredth Census: Whether Two Things Share a Root Is Decidable [Paper 300]. Zenodo. 10.5281/zenodo.22171914
- Yamagishi, Y. (2026). On Ice, Driving Slowly Can Be More Dangerous Than Driving Fast [Paper 453]. Zenodo. 10.5281/zenodo.22732924
Reference
Yamagishi, Y. (2026). Braking at Half Does Not Halve the Grip for Turning: Tyre Friction Shares One Circle Between Braking and Turning [Paper 593]. YOI: https://www.yamagishiyuuki.com/yoi/593
DOI: 10.5281/zenodo.23014398
ORCID: 0009-0006-8587-3450
PDF (English) · PDF (Japanese) · Date proof of the English PDF (OpenTimestamps .ots) · SHA256SUMS
The record page for this paper (in Japanese, with the PDFs in each language and the date proof): YOI 593. This paper is a preprint and has not been peer reviewed.